problem solving involving limits

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AP®︎/College Calculus AB

Course: ap®︎/college calculus ab   >   unit 1, strategy in finding limits.

• Conclusions from direct substitution (finding limits)
• Next steps after indeterminate form (finding limits)

Practice with direct substitution

• (Choice A)   The limit exists, and we found it! A The limit exists, and we found it!
• (Choice B)   The limit doesn't exist (probably an asymptote). B The limit doesn't exist (probably an asymptote).
• (Choice C)   The result is indeterminate. C The result is indeterminate.

Practice with the indeterminate form

• (Choice A)   Factorization and cancellation A Factorization and cancellation
• (Choice B)   Rationalization using conjugates B Rationalization using conjugates
• (Choice C)   Alternate forms of trigonometric functions C Alternate forms of trigonometric functions

Putting it all together

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2.3 The Limit Laws

Learning objectives.

• 2.3.1 Recognize the basic limit laws.
• 2.3.2 Use the limit laws to evaluate the limit of a function.
• 2.3.3 Evaluate the limit of a function by factoring.
• 2.3.4 Use the limit laws to evaluate the limit of a polynomial or rational function.
• 2.3.5 Evaluate the limit of a function by factoring or by using conjugates.
• 2.3.6 Evaluate the limit of a function by using the squeeze theorem.

In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.

Evaluating Limits with the Limit Laws

The first two limit laws were stated in Two Important Limits and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Theorem 2.4

Basic limit results.

For any real number a and any constant c ,

• lim x → a x = a lim x → a x = a (2.14)
• lim x → a c = c lim x → a c = c (2.15)

Example 2.13

Evaluating a basic limit.

Evaluate each of the following limits using Basic Limit Results .

• lim x → 2 x lim x → 2 x
• lim x → 2 5 lim x → 2 5
• The limit of x as x approaches a is a : lim x → 2 x = 2 . lim x → 2 x = 2 .
• The limit of a constant is that constant: lim x → 2 5 = 5 . lim x → 2 5 = 5 .

We now take a look at the limit laws , the individual properties of limits. The proofs that these laws hold are omitted here.

Theorem 2.5

Let f ( x ) f ( x ) and g ( x ) g ( x ) be defined for all x ≠ a x ≠ a over some open interval containing a . Assume that L and M are real numbers such that lim x → a f ( x ) = L lim x → a f ( x ) = L and lim x → a g ( x ) = M . lim x → a g ( x ) = M . Let c be a constant. Then, each of the following statements holds:

Sum law for limits : lim x → a ( f ( x ) + g ( x ) ) = lim x → a f ( x ) + lim x → a g ( x ) = L + M lim x → a ( f ( x ) + g ( x ) ) = lim x → a f ( x ) + lim x → a g ( x ) = L + M

Difference law for limits : lim x → a ( f ( x ) − g ( x ) ) = lim x → a f ( x ) − lim x → a g ( x ) = L − M lim x → a ( f ( x ) − g ( x ) ) = lim x → a f ( x ) − lim x → a g ( x ) = L − M

Constant multiple law for limits : lim x → a c f ( x ) = c · lim x → a f ( x ) = c L lim x → a c f ( x ) = c · lim x → a f ( x ) = c L

Product law for limits : lim x → a ( f ( x ) · g ( x ) ) = lim x → a f ( x ) · lim x → a g ( x ) = L · M lim x → a ( f ( x ) · g ( x ) ) = lim x → a f ( x ) · lim x → a g ( x ) = L · M

Quotient law for limits : lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) = L M lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) = L M for M ≠ 0 M ≠ 0

Power law for limits : lim x → a ( f ( x ) ) n = ( lim x → a f ( x ) ) n = L n lim x → a ( f ( x ) ) n = ( lim x → a f ( x ) ) n = L n for every positive integer n .

Root law for limits : lim x → a f ( x ) n = lim x → a f ( x ) n = L n lim x → a f ( x ) n = lim x → a f ( x ) n = L n for all L if n is odd and for L ≥ 0 L ≥ 0 if n is even and f ( x ) ≥ 0 f ( x ) ≥ 0 .

We now practice applying these limit laws to evaluate a limit.

Example 2.14

Evaluating a limit using limit laws.

Use the limit laws to evaluate lim x → −3 ( 4 x + 2 ) . lim x → −3 ( 4 x + 2 ) .

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

lim x → −3 ( 4 x + 2 ) = lim x → −3 4 x + lim x → −3 2 Apply the sum law. = 4 · lim x → −3 x + lim x → −3 2 Apply the constant multiple law. = 4 · ( −3 ) + 2 = −10 . Apply the basic limit results and simplify. lim x → −3 ( 4 x + 2 ) = lim x → −3 4 x + lim x → −3 2 Apply the sum law. = 4 · lim x → −3 x + lim x → −3 2 Apply the constant multiple law. = 4 · ( −3 ) + 2 = −10 . Apply the basic limit results and simplify.

Example 2.15

Using limit laws repeatedly.

Use the limit laws to evaluate lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 . lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 .

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 = lim x → 2 ( 2 x 2 − 3 x + 1 ) lim x → 2 ( x 3 + 4 ) Apply the quotient law, making sure that. ( 2 ) 3 + 4 ≠ 0 = 2 · lim x → 2 x 2 − 3 · lim x → 2 x + lim x → 2 1 lim x → 2 x 3 + lim x → 2 4 Apply the sum law and constant multiple law. = 2 · ( lim x → 2 x ) 2 − 3 · lim x → 2 x + lim x → 2 1 ( lim x → 2 x ) 3 + lim x → 2 4 Apply the power law. = 2 ( 4 ) − 3 ( 2 ) + 1 ( 2 ) 3 + 4 = 1 4 . Apply the basic limit laws and simplify. lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 = lim x → 2 ( 2 x 2 − 3 x + 1 ) lim x → 2 ( x 3 + 4 ) Apply the quotient law, making sure that. ( 2 ) 3 + 4 ≠ 0 = 2 · lim x → 2 x 2 − 3 · lim x → 2 x + lim x → 2 1 lim x → 2 x 3 + lim x → 2 4 Apply the sum law and constant multiple law. = 2 · ( lim x → 2 x ) 2 − 3 · lim x → 2 x + lim x → 2 1 ( lim x → 2 x ) 3 + lim x → 2 4 Apply the power law. = 2 ( 4 ) − 3 ( 2 ) + 1 ( 2 ) 3 + 4 = 1 4 . Apply the basic limit laws and simplify.

Checkpoint 2.11

Use the limit laws to evaluate lim x → 6 ( 2 x − 1 ) x + 4 . lim x → 6 ( 2 x − 1 ) x + 4 . In each step, indicate the limit law applied.

Limits of Polynomial and Rational Functions

By now you have probably noticed that, in each of the previous examples, it has been the case that lim x → a f ( x ) = f ( a ) . lim x → a f ( x ) = f ( a ) . This is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for which the rational function is defined.

Theorem 2.6

Let p ( x ) p ( x ) and q ( x ) q ( x ) be polynomial functions. Let a be a real number. Then,

To see that this theorem holds, consider the polynomial p ( x ) = c n x n + c n − 1 x n − 1 + ⋯ + c 1 x + c 0 . p ( x ) = c n x n + c n − 1 x n − 1 + ⋯ + c 1 x + c 0 . By applying the sum, constant multiple, and power laws, we end up with

It now follows from the quotient law that if p ( x ) p ( x ) and q ( x ) q ( x ) are polynomials for which q ( a ) ≠ 0 , q ( a ) ≠ 0 , then

Example 2.16 applies this result.

Example 2.16

Evaluating a limit of a rational function.

Evaluate the lim x → 3 2 x 2 − 3 x + 1 5 x + 4 . lim x → 3 2 x 2 − 3 x + 1 5 x + 4 .

Since 3 is in the domain of the rational function f ( x ) = 2 x 2 − 3 x + 1 5 x + 4 , f ( x ) = 2 x 2 − 3 x + 1 5 x + 4 , we can calculate the limit by substituting 3 for x into the function. Thus,

Checkpoint 2.12

Evaluate lim x → −2 ( 3 x 3 − 2 x + 7 ) . lim x → −2 ( 3 x 3 − 2 x + 7 ) .

Additional Limit Evaluation Techniques

As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for lim x → a f ( x ) lim x → a f ( x ) to exist when f ( a ) f ( a ) is undefined. The following observation allows us to evaluate many limits of this type:

If for all x ≠ a , f ( x ) = g ( x ) x ≠ a , f ( x ) = g ( x ) over some open interval containing a , then lim x → a f ( x ) = lim x → a g ( x ) . lim x → a f ( x ) = lim x → a g ( x ) .

To understand this idea better, consider the limit lim x → 1 x 2 − 1 x − 1 . lim x → 1 x 2 − 1 x − 1 .

The function

and the function g ( x ) = x + 1 g ( x ) = x + 1 are identical for all values of x ≠ 1 . x ≠ 1 . The graphs of these two functions are shown in Figure 2.24 .

We see that

The limit has the form lim x → a f ( x ) g ( x ) , lim x → a f ( x ) g ( x ) , where lim x → a f ( x ) = 0 lim x → a f ( x ) = 0 and lim x → a g ( x ) = 0 . lim x → a g ( x ) = 0 . (In this case, we say that f ( x ) / g ( x ) f ( x ) / g ( x ) has the indeterminate form 0 / 0 .) 0 / 0 .) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

Problem-Solving Strategy

Problem-solving strategy: calculating a limit when f ( x ) / g ( x ) f ( x ) / g ( x ) has the indeterminate form 0/0.

• First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.
• If f ( x ) f ( x ) and g ( x ) g ( x ) are polynomials, we should factor each function and cancel out any common factors.
• If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.
• If f ( x ) / g ( x ) f ( x ) / g ( x ) is a complex fraction, we begin by simplifying it.
• Last, we apply the limit laws.

The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-cancel technique; Example 2.18 shows multiplying by a conjugate. In Example 2.19 , we look at simplifying a complex fraction.

Example 2.17

Evaluating a limit by factoring and canceling.

Evaluate lim x → 3 x 2 − 3 x 2 x 2 − 5 x − 3 . lim x → 3 x 2 − 3 x 2 x 2 − 5 x − 3 .

Step 1. The function f ( x ) = x 2 − 3 x 2 x 2 − 5 x − 3 f ( x ) = x 2 − 3 x 2 x 2 − 5 x − 3 is undefined for x = 3 . x = 3 . In fact, if we substitute 3 into the function we get 0 / 0 , 0 / 0 , which is undefined. Factoring and canceling is a good strategy:

Step 2. For all x ≠ 3 , x 2 − 3 x 2 x 2 − 5 x − 3 = x 2 x + 1 . x ≠ 3 , x 2 − 3 x 2 x 2 − 5 x − 3 = x 2 x + 1 . Therefore,

Step 3. Evaluate using the limit laws:

Checkpoint 2.13

Evaluate lim x → −3 x 2 + 4 x + 3 x 2 − 9 . lim x → −3 x 2 + 4 x + 3 x 2 − 9 .

Example 2.18

Evaluating a limit by multiplying by a conjugate.

Evaluate lim x → −1 x + 2 − 1 x + 1 . lim x → −1 x + 2 − 1 x + 1 .

Step 1. x + 2 − 1 x + 1 x + 2 − 1 x + 1 has the form 0 / 0 0 / 0 at −1. Let’s begin by multiplying by x + 2 + 1 , x + 2 + 1 , the conjugate of x + 2 − 1 , x + 2 − 1 , on the numerator and denominator:

Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the ( x + 1 ) ( x + 1 ) in the denominator cancels out in the end:

Step 3. Then we cancel:

Step 4. Last, we apply the limit laws:

Checkpoint 2.14

Evaluate lim x → 5 x − 1 − 2 x − 5 . lim x → 5 x − 1 − 2 x − 5 .

Example 2.19

Evaluating a limit by simplifying a complex fraction.

Evaluate lim x → 1 1 x + 1 − 1 2 x − 1 . lim x → 1 1 x + 1 − 1 2 x − 1 .

Step 1. 1 x + 1 − 1 2 x − 1 1 x + 1 − 1 2 x − 1 has the form 0 / 0 0 / 0 at 1. We simplify the algebraic fraction by multiplying by 2 ( x + 1 ) / 2 ( x + 1 ) : 2 ( x + 1 ) / 2 ( x + 1 ) :

Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor ( x − 1 ) : ( x − 1 ) :

Step 3. Then, we simplify the numerator:

Step 4. Now we factor out −1 from the numerator:

Step 5. Then, we cancel the common factors of ( x − 1 ) : ( x − 1 ) :

Step 6. Last, we evaluate using the limit laws:

Checkpoint 2.15

Evaluate lim x → −3 1 x + 2 + 1 x + 3 . lim x → −3 1 x + 2 + 1 x + 3 .

Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.

Example 2.20

Evaluating a limit when the limit laws do not apply.

Evaluate lim x → 0 ( 1 x + 5 x ( x − 5 ) ) . lim x → 0 ( 1 x + 5 x ( x − 5 ) ) .

Both 1 / x 1 / x and 5 / x ( x − 5 ) 5 / x ( x − 5 ) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that

Checkpoint 2.16

Evaluate lim x → 3 ( 1 x − 3 − 4 x 2 − 2 x − 3 ) . lim x → 3 ( 1 x − 3 − 4 x 2 − 2 x − 3 ) .

Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form lim x → a − h ( x ) , lim x → a − h ( x ) , we require the function h ( x ) h ( x ) to be defined over an open interval of the form ( b , a ) ; ( b , a ) ; for a limit of the form lim x → a + h ( x ) , lim x → a + h ( x ) , we require the function h ( x ) h ( x ) to be defined over an open interval of the form ( a , c ) . ( a , c ) . Example 2.21 illustrates this point.

Example 2.21

Evaluating a one-sided limit using the limit laws.

Evaluate each of the following limits, if possible.

• lim x → 3 − x − 3 lim x → 3 − x − 3
• lim x → 3 + x − 3 lim x → 3 + x − 3

Figure 2.25 illustrates the function f ( x ) = x − 3 f ( x ) = x − 3 and aids in our understanding of these limits.

• The function f ( x ) = x − 3 f ( x ) = x − 3 is defined over the interval [ 3 , + ∞ ) . [ 3 , + ∞ ) . Since this function is not defined to the left of 3, we cannot apply the limit laws to compute lim x → 3 − x − 3 . lim x → 3 − x − 3 . In fact, since f ( x ) = x − 3 f ( x ) = x − 3 is undefined to the left of 3, lim x → 3 − x − 3 lim x → 3 − x − 3 does not exist.
• Since f ( x ) = x − 3 f ( x ) = x − 3 is defined to the right of 3, the limit laws do apply to lim x → 3 + x − 3 . lim x → 3 + x − 3 . By applying these limit laws we obtain lim x → 3 + x − 3 = 0 . lim x → 3 + x − 3 = 0 .

In Example 2.22 we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

Example 2.22

Evaluating a two-sided limit using the limit laws.

For f ( x ) = { 4 x − 3 if x < 2 ( x − 3 ) 2 if x ≥ 2 , f ( x ) = { 4 x − 3 if x < 2 ( x − 3 ) 2 if x ≥ 2 , evaluate each of the following limits:

• lim x → 2 − f ( x ) lim x → 2 − f ( x )
• lim x → 2 + f ( x ) lim x → 2 + f ( x )
• lim x → 2 f ( x ) lim x → 2 f ( x )

Figure 2.26 illustrates the function f ( x ) f ( x ) and aids in our understanding of these limits.

• Since f ( x ) = 4 x − 3 f ( x ) = 4 x − 3 for all x in ( − ∞ , 2 ) , ( − ∞ , 2 ) , replace f ( x ) f ( x ) in the limit with 4 x − 3 4 x − 3 and apply the limit laws: lim x → 2 − f ( x ) = lim x → 2 − ( 4 x − 3 ) = 5 . lim x → 2 − f ( x ) = lim x → 2 − ( 4 x − 3 ) = 5 .
• Since f ( x ) = ( x − 3 ) 2 f ( x ) = ( x − 3 ) 2 for all x in ( 2 , + ∞ ) , ( 2 , + ∞ ) , replace f ( x ) f ( x ) in the limit with ( x − 3 ) 2 ( x − 3 ) 2 and apply the limit laws: lim x → 2 + f ( x ) = lim x → 2 + ( x − 3 ) 2 = 1 . lim x → 2 + f ( x ) = lim x → 2 + ( x − 3 ) 2 = 1 .
• Since lim x → 2 − f ( x ) = 5 lim x → 2 − f ( x ) = 5 and lim x → 2 + f ( x ) = 1 , lim x → 2 + f ( x ) = 1 , we conclude that lim x → 2 f ( x ) lim x → 2 f ( x ) does not exist.

Checkpoint 2.17

Graph f ( x ) = { − x − 2 if x < − 1 2 if x = −1 x 3 if x > − 1 f ( x ) = { − x − 2 if x < − 1 2 if x = −1 x 3 if x > − 1 and evaluate lim x → −1 − f ( x ) . lim x → −1 − f ( x ) .

We now turn our attention to evaluating a limit of the form lim x → a f ( x ) g ( x ) , lim x → a f ( x ) g ( x ) , where lim x → a f ( x ) = K , lim x → a f ( x ) = K , where K ≠ 0 K ≠ 0 and lim x → a g ( x ) = 0 . lim x → a g ( x ) = 0 . That is, f ( x ) / g ( x ) f ( x ) / g ( x ) has the form K / 0 , K ≠ 0 K / 0 , K ≠ 0 at a .

Example 2.23

Evaluating a limit of the form k / 0 , k ≠ 0 k / 0 , k ≠ 0 using the limit laws.

Evaluate lim x → 2 − x − 3 x 2 − 2 x . lim x → 2 − x − 3 x 2 − 2 x .

Step 1. After substituting in x = 2 , x = 2 , we see that this limit has the form −1 / 0 . −1 / 0 . That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of x − 3 x ( x − 2 ) x − 3 x ( x − 2 ) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

Step 2. Since x − 2 x − 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1 / ( x − 2 ) 1 / ( x − 2 ) from the rest of the function:

Step 3. lim x → 2 − x − 3 x = − 1 2 lim x → 2 − x − 3 x = − 1 2 and lim x → 2 − 1 x − 2 = − ∞ . lim x → 2 − 1 x − 2 = − ∞ . Therefore, the product of ( x − 3 ) / x ( x − 3 ) / x and 1 / ( x − 2 ) 1 / ( x − 2 ) has a limit of +∞: +∞:

Checkpoint 2.18

Evaluate lim x → 1 x + 2 ( x − 1 ) 2 . lim x → 1 x + 2 ( x − 1 ) 2 .

The Squeeze Theorem

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem , proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a . Figure 2.27 illustrates this idea.

Theorem 2.7

Let f ( x ) , g ( x ) , f ( x ) , g ( x ) , and h ( x ) h ( x ) be defined for all x ≠ a x ≠ a over an open interval containing a . If

for all x ≠ a x ≠ a in an open interval containing a and

where L is a real number, then lim x → a g ( x ) = L . lim x → a g ( x ) = L .

Example 2.24

Applying the squeeze theorem.

Apply the squeeze theorem to evaluate lim x → 0 x cos x . lim x → 0 x cos x .

Because −1 ≤ cos x ≤ 1 −1 ≤ cos x ≤ 1 for all x , we have − | x | ≤ x cos x ≤ | x | − | x | ≤ x cos x ≤ | x | . Since lim x → 0 ( − | x | ) = 0 = lim x → 0 | x | , lim x → 0 ( − | x | ) = 0 = lim x → 0 | x | , from the squeeze theorem, we obtain lim x → 0 x cos x = 0 . lim x → 0 x cos x = 0 . The graphs of f ( x ) = − | x | , g ( x ) = x cos x , f ( x ) = − | x | , g ( x ) = x cos x , and h ( x ) = | x | h ( x ) = | x | are shown in Figure 2.28 .

Checkpoint 2.19

Use the squeeze theorem to evaluate lim x → 0 x 2 sin 1 x . lim x → 0 x 2 sin 1 x .

We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is lim θ → 0 sin θ . lim θ → 0 sin θ . Consider the unit circle shown in Figure 2.29 . In the figure, we see that sin θ sin θ is the y -coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for 0 < θ < π 2 , 0 < sin θ < θ . 0 < θ < π 2 , 0 < sin θ < θ .

Because lim θ → 0 + 0 = 0 lim θ → 0 + 0 = 0 and lim θ → 0 + θ = 0 , lim θ → 0 + θ = 0 , by using the squeeze theorem we conclude that

To see that lim θ → 0 − sin θ = 0 lim θ → 0 − sin θ = 0 as well, observe that for − π 2 < θ < 0 , 0 < − θ < π 2 − π 2 < θ < 0 , 0 < − θ < π 2 and hence, 0 < sin ( − θ ) < − θ . 0 < sin ( − θ ) < − θ . Consequently, 0 < − sin θ < − θ . 0 < − sin θ < − θ . It follows that 0 > sin θ > θ . 0 > sin θ > θ . An application of the squeeze theorem produces the desired limit. Thus, since lim θ → 0 + sin θ = 0 lim θ → 0 + sin θ = 0 and lim θ → 0 − sin θ = 0 , lim θ → 0 − sin θ = 0 ,

Next, using the identity cos θ = 1 − sin 2 θ cos θ = 1 − sin 2 θ for − π 2 < θ < π 2 , − π 2 < θ < π 2 , we see that

We now take a look at a limit that plays an important role in later chapters—namely, lim θ → 0 sin θ θ . lim θ → 0 sin θ θ . To evaluate this limit, we use the unit circle in Figure 2.30 . Notice that this figure adds one additional triangle to Figure 2.30 . We see that the length of the side opposite angle θ in this new triangle is tan θ . tan θ . Thus, we see that for 0 < θ < π 2 , sin θ < θ < tan θ . 0 < θ < π 2 , sin θ < θ < tan θ .

By dividing by sin θ sin θ in all parts of the inequality, we obtain

Equivalently, we have

Since lim θ → 0 + 1 = 1 = lim θ → 0 + cos θ , lim θ → 0 + 1 = 1 = lim θ → 0 + cos θ , we conclude that lim θ → 0 + sin θ θ = 1 . lim θ → 0 + sin θ θ = 1 . By applying a manipulation similar to that used in demonstrating that lim θ → 0 − sin θ = 0 , lim θ → 0 − sin θ = 0 , we can show that lim θ → 0 − sin θ θ = 1 . lim θ → 0 − sin θ θ = 1 . Thus,

In Example 2.25 we use this limit to establish lim θ → 0 1 − cos θ θ = 0 . lim θ → 0 1 − cos θ θ = 0 . This limit also proves useful in later chapters.

Example 2.25

Evaluating an important trigonometric limit.

Evaluate lim θ → 0 1 − cos θ θ . lim θ → 0 1 − cos θ θ .

In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

Checkpoint 2.20

Evaluate lim θ → 0 1 − cos θ sin θ . lim θ → 0 1 − cos θ sin θ .

Student Project

Deriving the formula for the area of a circle.

Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the methods of calculus. The Greek mathematician Archimedes (ca. 287−212; BCE) was particularly inventive, using polygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased. He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could have predicted about the limit.

We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regular polygon as being made up of n triangles. By taking the limit as the vertex angle of these triangles goes to zero, you can obtain the area of the circle. To see this, carry out the following steps:

• Using the expressions that you obtained in step 1, express the area of the isosceles triangle in terms of θ and r . (Substitute ( 1 / 2 ) sin θ ( 1 / 2 ) sin θ for sin ( θ / 2 ) cos ( θ / 2 ) sin ( θ / 2 ) cos ( θ / 2 ) in your expression.)
• If an n -sided regular polygon is inscribed in a circle of radius r , find a relationship between θ and n . Solve this for n . Keep in mind there are 2 π radians in a circle. (Use radians, not degrees.)
• Find an expression for the area of the n -sided polygon in terms of r and θ .
• To find a formula for the area of the circle, find the limit of the expression in step 4 as θ goes to zero. ( Hint: lim θ → 0 ( sin θ ) θ = 1 ). lim θ → 0 ( sin θ ) θ = 1 ).

The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration .

Section 2.3 Exercises

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

lim x → 0 ( 4 x 2 − 2 x + 3 ) lim x → 0 ( 4 x 2 − 2 x + 3 )

lim x → 1 x 3 + 3 x 2 + 5 4 − 7 x lim x → 1 x 3 + 3 x 2 + 5 4 − 7 x

lim x → −2 x 2 − 6 x + 3 lim x → −2 x 2 − 6 x + 3

lim x → −1 ( 9 x + 1 ) 2 lim x → −1 ( 9 x + 1 ) 2

In the following exercises, use direct substitution to evaluate each limit.

lim x → 7 x 2 lim x → 7 x 2

lim x → −2 ( 4 x 2 − 1 ) lim x → −2 ( 4 x 2 − 1 )

lim x → 0 1 1 + sin x lim x → 0 1 1 + sin x

lim x → 2 e 2 x − x 2 lim x → 2 e 2 x − x 2

lim x → 1 2 − 7 x x + 6 lim x → 1 2 − 7 x x + 6

lim x → 3 ln e 3 x lim x → 3 ln e 3 x

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0 / 0 . 0 / 0 . Then, evaluate the limit.

lim x → 4 x 2 − 16 x − 4 lim x → 4 x 2 − 16 x − 4

lim x → 2 x − 2 x 2 − 2 x lim x → 2 x − 2 x 2 − 2 x

lim x → 6 3 x − 18 2 x − 12 lim x → 6 3 x − 18 2 x − 12

lim h → 0 ( 1 + h ) 2 − 1 h lim h → 0 ( 1 + h ) 2 − 1 h

lim t → 9 t − 9 t − 3 lim t → 9 t − 9 t − 3

lim h → 0 1 a + h − 1 a h , lim h → 0 1 a + h − 1 a h , where a is a non-zero real-valued constant

lim θ → π sin θ tan θ lim θ → π sin θ tan θ

lim x → 1 x 3 − 1 x 2 − 1 lim x → 1 x 3 − 1 x 2 − 1

lim x → 1 / 2 2 x 2 + 3 x − 2 2 x − 1 lim x → 1 / 2 2 x 2 + 3 x − 2 2 x − 1

lim x → −3 x + 4 − 1 x + 3 lim x → −3 x + 4 − 1 x + 3

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.23 to simplify the function to help determine the limit.

lim x → −2 − 2 x 2 + 7 x − 4 x 2 + x − 2 lim x → −2 − 2 x 2 + 7 x − 4 x 2 + x − 2

lim x → −2 + 2 x 2 + 7 x − 4 x 2 + x − 2 lim x → −2 + 2 x 2 + 7 x − 4 x 2 + x − 2

lim x → 1 − 2 x 2 + 7 x − 4 x 2 + x − 2 lim x → 1 − 2 x 2 + 7 x − 4 x 2 + x − 2

lim x → 1 + 2 x 2 + 7 x − 4 x 2 + x − 2 lim x → 1 + 2 x 2 + 7 x − 4 x 2 + x − 2

In the following exercises, assume that lim x → 6 f ( x ) = 4 , lim x → 6 g ( x ) = 9 , lim x → 6 f ( x ) = 4 , lim x → 6 g ( x ) = 9 , and lim x → 6 h ( x ) = 6 . lim x → 6 h ( x ) = 6 . Use these three facts and the limit laws to evaluate each limit.

lim x → 6 2 f ( x ) g ( x ) lim x → 6 2 f ( x ) g ( x )

lim x → 6 g ( x ) − 1 f ( x ) lim x → 6 g ( x ) − 1 f ( x )

lim x → 6 ( f ( x ) + 1 3 g ( x ) ) lim x → 6 ( f ( x ) + 1 3 g ( x ) )

lim x → 6 ( h ( x ) ) 3 2 lim x → 6 ( h ( x ) ) 3 2

lim x → 6 g ( x ) − f ( x ) lim x → 6 g ( x ) − f ( x )

lim x → 6 x · h ( x ) lim x → 6 x · h ( x )

lim x → 6 [ ( x + 1 ) · f ( x ) ] lim x → 6 [ ( x + 1 ) · f ( x ) ]

lim x → 6 ( f ( x ) · g ( x ) − h ( x ) ) lim x → 6 ( f ( x ) · g ( x ) − h ( x ) )

[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

f ( x ) = { x 2 , x ≤ 3 x + 4 , x > 3 f ( x ) = { x 2 , x ≤ 3 x + 4 , x > 3

• lim x → 3 − f ( x ) lim x → 3 − f ( x )
• lim x → 3 + f ( x ) lim x → 3 + f ( x )

g ( x ) = { x 3 − 1 , x ≤ 0 1 , x > 0 g ( x ) = { x 3 − 1 , x ≤ 0 1 , x > 0

• lim x → 0 − g ( x ) lim x → 0 − g ( x )
• lim x → 0 + g ( x ) lim x → 0 + g ( x )

h ( x ) = { x 2 − 2 x + 1 , x < 2 3 − x , x ≥ 2 h ( x ) = { x 2 − 2 x + 1 , x < 2 3 − x , x ≥ 2

• lim x → 2 − h ( x ) lim x → 2 − h ( x )
• lim x → 2 + h ( x ) lim x → 2 + h ( x )

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

lim x → −3 + ( f ( x ) + g ( x ) ) lim x → −3 + ( f ( x ) + g ( x ) )

lim x → −3 − ( f ( x ) − 3 g ( x ) ) lim x → −3 − ( f ( x ) − 3 g ( x ) )

lim x → 0 f ( x ) g ( x ) 3 lim x → 0 f ( x ) g ( x ) 3

lim x → −5 2 + g ( x ) f ( x ) lim x → −5 2 + g ( x ) f ( x )

lim x → 1 ( f ( x ) ) 2 lim x → 1 ( f ( x ) ) 2

lim x → 1 f ( x ) − g ( x ) 3 lim x → 1 f ( x ) − g ( x ) 3

lim x → −7 ( x · g ( x ) ) lim x → −7 ( x · g ( x ) )

lim x → −9 [ x · f ( x ) + 2 · g ( x ) ] lim x → −9 [ x · f ( x ) + 2 · g ( x ) ]

For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions f ( x ) , g ( x ) , f ( x ) , g ( x ) , and h ( x ) h ( x ) when possible.

[T] True or False? If 2 x − 1 ≤ g ( x ) ≤ x 2 − 2 x + 3 , 2 x − 1 ≤ g ( x ) ≤ x 2 − 2 x + 3 , then lim x → 2 g ( x ) = 0 . lim x → 2 g ( x ) = 0 .

[T] lim θ → 0 θ 2 cos ( 1 θ ) lim θ → 0 θ 2 cos ( 1 θ )

lim x → 0 f ( x ) , lim x → 0 f ( x ) , where f ( x ) = { 0 , x rational x 2 , x irrational f ( x ) = { 0 , x rational x 2 , x irrational

[T] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: E ( r ) = q 4 π ε 0 r 2 , E ( r ) = q 4 π ε 0 r 2 , where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and 1 4 π ε 0 1 4 π ε 0 is Coulomb’s constant: 8.988 × 10 9 N · m 2 / C 2 . 8.988 × 10 9 N · m 2 / C 2 .

• Use a graphing calculator to graph E ( r ) E ( r ) given that the charge of the particle is q = 10 −10 . q = 10 −10 .
• Evaluate lim r → 0 + E ( r ) . lim r → 0 + E ( r ) . What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

[T] The density of an object is given by its mass divided by its volume: ρ = m / V . ρ = m / V .

• Use a calculator to plot the volume as a function of density ( V = m / ρ ) , ( V = m / ρ ) , assuming you are examining something of mass 8 kg ( m = 8 ). m = 8 ).
• Evaluate lim ρ → 0 + V ( ρ ) lim ρ → 0 + V ( ρ ) and explain the physical meaning.

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• Section URL: https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws

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Evaluating Limits - Limits Practice Problems

Problem: evaluating limits #1, evaluate the limit: \large{\lim_{x\to 3}\frac{5x^2-3x}{3x^3+3}}.

The first step is to always try solving the limit by direct substitution. So, we will substitute 3 into the function and see what we get.

This was a simple problem which involved evaluating limits by direct substitution!

Solution Complete!

Problem: Evaluating Limits #2

Evaluate the limit: \large{\lim_{x\to 0}\frac{3x^2-2x}{x}}.

First, we will try and substitute the limit into the function. So, substituting the 0 into the function, we have:

So, we end up with \frac{0}{0} which is an indeterminate form.

Since we ended up with an indeterminate form, we need to rewrite the original function in another way, and then try to solve the limit again.

Notice how we can factor out the x from the numerator, as follows:

As you see, with some algebraic manipulation, we were able to rewrite the original function.

Now we can simply plug in the limit and solve, as follows:

Problem: Evaluating Limits #3

Evaluate the limit: \large{\lim_{h\to -2}\frac{h^2+3h+2}{h+2}}.

First, substituting the limit directly into the function, gives us:

Hence, we have an indeterminate form, which means we must rewrite the function.

Solution Locked! Click to View!

Problem: Evaluating Limits #4

Evaluate the limit: \large{\lim_{x\to \frac{\pi}{2}}\frac{sin(x)}{cos(x)-1}}.

Let’s attempt to solve the limit by direct substitution. We have:

Problem: Evaluating Limits #5

Evaluate the limit: \large{\lim_{k\to 1}\frac{k-\sqrt{k}}{k^2-1}}.

First, let’s try substituting the limit into the function, as shown:

So, we end up with an indeterminate form. Thus, we must rewrite the function in some way before  plugging in the limit.

When we are rewriting functions that involve roots, we want to do what is known as rationalizing the denominator. We do this by multiplying the function with it’s conjugate, as follows:

Problem: Evaluating Limits #6

Evaluate the limit: \large{\lim_{x\to -2}\frac{x^2-4}{x+2}}.

If we substitute the limit into the function, we get:

Of course, this is an indeterminate form, so we must change the function in some way, before substituting the limit.

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• Trigonometry

Limits Questions with Solutions

• Math Doubts

Once you are confident about the limit rules , you are ready to use them in the limits problems. The list of questions on limits with answers is given here for your practice. A worksheet with limits examples and solutions for you to learn how to evaluate the limits of the functions by the limits formulas in calculus.

Limits methods

Direct substitution, factorization.

$(1).\,\,$ Evaluate $\displaystyle \large \lim_{x\,\to\,5}{\normalsize \dfrac{x^2-25}{x-5}}$

$(2).\,\,$ Evaluate $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3+7x^2-36}{x^2+2x-8}}$

List of limits questions on factorisation to learn how to find the limit of a function by factoring.

Rationalization

$(1).\,\,$ Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt{1+x}-1}{x}}$

$(2).\,\,$ Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{x}{\sqrt{a+x}-\sqrt{a-x}}}$

List of limit questions on rationalisation with solutions to learn how to find the limits by rationalization.

L’Hôpital’s Rule

$(1).\,\,$ Evaluate $\displaystyle \large \lim_{x\,\to\,-3}{\normalsize \dfrac{6+2x}{x^2+3x}}$

$(2).\,\,$ Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^x-1-x}{x^2}}$

Limit of functions

Algebraic functions.

List of limit problems with solutions for the algebraic functions to find the limits of functions in which algebraic functions are involved.

Trigonometric functions

Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$

List of limit problems with solutions for the trigonometric functions to find the limits of functions in which trigonometric functions are involved.

Logarithmic functions

List of limit problems with solutions for the logarithmic functions to find the limits of functions in which logarithmic functions are involved.

Exponential functions

List of limit problems with solutions for the exponential functions to evaluate the limits of functions in which exponential functions are involved.

Difficult Problems

List of hard/tough problems in limits with solutions.

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1.3: Finding Limits Analytically

• Last updated
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• Page ID 4152

• Gregory Hartman et al.
• Virginia Military Institute

In Section 1.1 we explored the concept of the limit without a strict definition, meaning we could only make approximations. In the previous section we gave the definition of the limit and demonstrated how to use it to verify our approximations were correct. Thus far, our method of finding a limit is 1) make a really good approximation either graphically or numerically, and 2) verify our approximation is correct using a \(\epsilon\)-\(\delta\) proof.

This process has its shortcomings, not the least of which is the fact that \(\epsilon\)--\(\delta\) proofs are cumbersome. This section gives a series of theorems which allow us to find limits much more quickly and intuitively.

Recognizing that \(\epsilon\)-\(\delta\) proofs are cumbersome, this section gives a series of theorems which allow us to find limits much more quickly and intuitively.

Suppose that \(\lim\limits_{x\to 2} f(x)=2\) and \(\lim\limits_{x\to 2} g(x) = 3\).What is \(\lim\limits_{x\to 2}(f(x)+g(x))\)? Intuition tells us that the limit should be 5, as we expect limits to behave in a nice way. The following theorem states that already established limits do behave nicely.

Theorem \(\PageIndex{1}\): Basic Limit Properties

Let \(b\), \(c\), \(L\) and \(K\) be real numbers, let \(n\) be a positive integer, and let \(f\) and \(g\) be functions with the following limits:

\[\lim\limits_{x\to c}f(x) = L \text{ and } \lim\limits_{x\to c} g(x) = K.\nonumber\]

The following limits hold.

• Constants: \(\displaystyle \lim\limits_{x\to c} b = b\)
• Identity: \(\displaystyle \lim\limits_{x\to c} x = c\)
• Sums/Differences: \(\displaystyle \lim\limits_{x\to c}(f(x)\pm g(x)) = L\pm K\)
• Scalar Multiples: \(\displaystyle \lim\limits_{x\to c} b\cdot f(x) = bL\)
• Products: \(\displaystyle \lim\limits_{x\to c} f(x)\cdot g(x) = LK\)
• Quotients: \(\displaystyle \lim\limits_{x\to c} f(x)/g(x) = L/K\), (\(K\neq 0)\)
• Powers: \(\displaystyle \lim\limits_{x\to c} f(x)^n = L^n\)
• Roots: \(\displaystyle \lim\limits_{x\to c} \sqrt[n]{f(x)} = \sqrt[n]{L}\)
• Compositions: Adjust our previously given limit situation to:

\[\lim\limits_{x\to c}f(x) = L,\ \lim\limits_{x\to L} g(x) = K \text{ and } g(L)=K .\]\[\text{Then } \lim\limits_{x\to c}g(f(x)) = K.\]

We make a note about Property #8: when \(n\) is even, \(L\) must be greater than 0. If \(n\) is odd, then the statement is true for all \(L\).

We apply the theorem to an example.

Example \(\PageIndex{1}\): Using Basic Limit Properties

\[\lim\limits_{x\to 2} f(x)=2,\quad\lim\limits_{x\to 2} g(x) = 3\quad \text{ and }\quad p(x) = 3x^2-5x+7.\]

Find the following limits:

• \( \lim\limits_{x\to 2} \big(f(x) + g(x)\big)\)
• \( \lim\limits_{x\to 2} \big(5f(x) + g(x)^2\big)\)
• \( \lim\limits_{x\to 2} p(x)\)
• Using the Sum/Difference rule, we know that \( \lim\limits_{x\to 2} \big(f(x) + g(x)\big) = 2+3 =5\).
• Using the Scalar Multiple and Sum/Difference rules, we find that \( \lim\limits_{x\to 2} \big(5f(x) + g(x)^2\big) = 5\cdot 2 + 3^2 = 19.\)
• Here we combine the Power, Scalar Multiple, Sum/Difference and Constant Rules. We show quite a few steps, but in general these can be omitted:

\[\begin{align*} \lim\limits_{x\to 2} p(x) &= \lim\limits_{x\to 2} (3x^2-5x+7) \\ &= \lim\limits_{x\to 2} 3x^2-\lim\limits_{x\to 2} 5x+\lim\limits_{x\to 2}7 \\ &= 3\cdot 2^2 - 5\cdot 2+7 \\ &= 9 \end{align*}\]

Part 3 of the previous example demonstrates how the limit of a quadratic polynomial can be determined using the properties of Theorem 1. Not only that, recognize that

\[\lim\limits_{x\to 2} p(x) = 9 = p(2);\]

i.e., the limit at 2 was found just by plugging 2 into the function. This holds true for all polynomials, and also for rational functions (which are quotients of polynomials), as stated in the following theorem.

Theorem \(\PageIndex{2}\) : Limits of Polynomial and Rational Functions

Limits of Polynomial and Rational Functions}{Let \(p(x)\) and \(q(x)\) be polynomials and \(c\) a real number. Then:

• \( \lim\limits_{x\to c} p(x) = p(c)\)
• \( \lim\limits_{x\to c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}\), where \(q(c) \neq 0\).

Example \(\PageIndex{2}\): Finding a limit of a rational function

Using Theorem 2, find

\[\lim\limits_{x\to -1} \frac{3x^2-5x+1}{x^4-x^2+3}. \nonumber\]

Using Theorem \(\PageIndex{2}\), we can quickly state that

\[\begin{align*} \lim\limits_{x\to -1}\frac{3x^2-5x+1}{x^4-x^2+3} &= \frac{3(-1)^2-5(-1)+1}{(-1)^4-(-1)^2+3} \\ &= \frac{9}{3} =3. \end{align*}\]

It was likely frustrating in Section 1.2 to do a lot of work to prove that

\[\lim\limits_{x\to 2} x^2 = 4\]

as it seemed fairly obvious. The previous theorems state that many functions behave in such an "obvious'' fashion, as demonstrated by the rational function in Example 11.

Polynomial and rational functions are not the only functions to behave in such a predictable way. The following theorem gives a list of functions whose behavior is particularly "nice'' in terms of limits. In the next section, we will give a formal name to these functions that behave "nicely.''

Theorem \(\PageIndex{4}\): Special Limits

Let \(c\) be a real number in the domain of the given function and let \(n\) be a positive integer. The following limits hold:

\[\begin{align} &1.\lim\limits_{x\to c} \sin x = \sin c \quad &&4. \lim\limits_{x\to c} \csc x = \csc c\quad &&&7.\lim\limits_{x\to c} a^x = a^c (a>0) \\ &2. \lim\limits_{x\to c} \cos x = \cos c \quad &&5. \lim\limits_{x\to c} \sec x = \sec c \quad &&&8.\lim\limits_{x\to c} \ln x = \ln c \\ &3. \lim\limits_{x\to c} \tan x = \tan c \quad &&6.\lim\limits_{x\to c} \cot x = \cot c \quad &&&9. \lim\limits_{x\to c} \sqrt[n]{x} = \sqrt[n]{c} \end{align}\]

Example \(\PageIndex{3}\): Evaluating limits analytically

Evaluate the following limits.

• \(\lim\limits_{x\to \pi} \cos x\)
• \( \lim\limits_{x\to 3} (\sec^2x - \tan^2 x)\)
• \( \lim\limits_{x\to \pi/2} \cos x\sin x\)
• \( \lim\limits_{x\to 1} e^{\ln x}\)
• \( \lim\limits_{x\to 0} \frac{\sin x}{x}\)
• This is a straightforward application of Theorem 3. \(\lim\limits_{x\to \pi} \cos x = \cos \pi = -1\).
• We can approach this in at least two ways. First, by directly applying Theorem 3, we have:\[\lim\limits_{x\to 3} (\sec^2x - \tan^2 x) = \sec^23-\tan^23.\]Using the Pythagorean Theorem, this last expression is 1; therefore \[\lim\limits_{x\to 3} (\sec^2x - \tan^2 x) = 1.\]We can also use the Pythagorean Theorem from the start. \[\lim\limits_{x\to 3} (\sec^2x - \tan^2 x) = \lim\limits_{x\to 3} 1 = 1,\]using the Constant limit rule. Either way, we find the limit is 1.
• Applying the Product limit rule of Theorem 1 and Theorem 3 gives \[ \lim\limits_{x\to \pi/2} \cos x\sin x = \cos (\pi/2)\sin(\pi/2) = 0\cdot 1 = 0.\]
• Again, we can approach this in two ways. First, we can use the exponential/logarithmic identity that \(e^{\ln x} = x\) and evaluate \( \lim\limits_{x\to 1} e^{\ln x} = \lim\limits_{x\to 1} x = 1.\) We can also use the limit Composition Rule of Theorem 1. Using Theorem 3, we have \( \lim\limits_{x\to 1}\ln x = \ln 1 = 0\) and \(\lim\limits_{x\to 0} e^x= e^0=1\), satisfying the conditions of the Composition Rule. Applying this rule, \[ \lim\limits_{x\to 1} e^{\ln x} = \lim\limits_{x\to 0} e^x = e^0 = 1.\]Both approaches are valid, giving the same result.
• We encountered this limit in Section 1.1. Applying our theorems, we attempt to find the limit as \[\lim\limits_{x\to 0}\frac{\sin x}{x}\rightarrow \frac{\sin 0}{0} \rightarrow \frac{"\,0\,"}{0}.\]This, of course, violates a condition of Theorem 1, as the limit of the denominator is not allowed to be 0. Therefore, we are still unable to evaluate this limit with tools we currently have at hand.

The section could have been titled "Using Known Limits to Find Unknown Limits.'' By knowing certain limits of functions, we can find limits involving sums, products, powers, etc., of these functions. We further the development of such comparative tools with the Squeeze Theorem, a clever and intuitive way to find the value of some limits.

Before stating this theorem formally, suppose we have functions \(f\), \(g\) and \(h\) where \(g\) always takes on values between \(f\) and \(h\); that is, for all \(x\) in an interval,

\[f(x) \leq g(x) \leq h(x).\]

If \(f\) and \(h\) have the same limit at \(c\), and \(g\) is always "squeezed'' between them, then \(g\) must have the same limit as well. That is what the Squeeze Theorem states.

Theorem \(\PageIndex{5}\): Squeeze Theorem

Let \(f\), \(g\) and \(h\) be functions on an open interval \(I\) containing \(c\) such that for all \(x\) in \(I\),

\[f(x)\leq g(x) \leq h(x).\]

\[\lim\limits_{x\to c} f(x) = L = \lim\limits_{x\to c} h(x),\]

\[\lim\limits_{x\to c} g(x) = L.\]

It can take some work to figure out appropriate functions by which to "squeeze'' the given function of which you are trying to evaluate a limit. However, that is generally the only place work is necessary; the theorem makes the "evaluating the limit part'' very simple.

We use the Squeeze Theorem in the following example to finally prove that \( \lim\limits_{x\to 0} \frac{\sin x}{x} = 1\).

Example \(\PageIndex{4}\): Using the Squeeze Theorem

Use the Squeeze Theorem to show that

\[ \lim\limits_{x\to 0} \frac{\sin x}{x} = 1. \nonumber\]

We begin by considering the unit circle. Each point on the unit circle has coordinates \((\cos \theta,\sin \theta)\) for some angle \(\theta\) as shown in Figure \(\PageIndex{1}\). Using similar triangles, we can extend the line from the origin through the point to the point \((1,\tan \theta)\), as shown. (Here we are assuming that \(0\leq \theta \leq \pi/2\).Later we will show that we can also consider \(\theta \leq 0\).)

Figure 1.19 shows three regions have been constructed in the first quadrant, two triangles and a sector of a circle, which are also drawn below. The area of the large triangle is \(\frac12\tan\theta\); the area of the sector is \(\theta/2\); the area of the triangle contained inside the sector is \(\frac12\sin\theta\). It is then clear from the diagram that

Multiply all terms by \(\frac{2}{\sin \theta}\), giving

\[\frac{1}{\cos\theta} \geq \frac{\theta}{\sin \theta} \geq 1.\]

Taking reciprocals reverses the inequalities, giving

\[ \cos \theta \leq \frac{\sin \theta}{\theta} \leq 1.\]

(These inequalities hold for all values of \(\theta\) near 0, even negative values, since \(\cos (-\theta) = \cos \theta\) and \(\sin (-\theta) = -\sin \theta\).)

Now take limits.

\[\lim\limits_{\theta\to 0} \cos \theta \leq \lim\limits_{\theta\to 0} \frac{\sin\theta}{\theta} \leq \lim\limits_{\theta\to 0} 1 \]

\[\cos 0 \leq \lim\limits_{\theta\to 0} \frac{\sin\theta}{\theta} \leq 1 \]

\[1 \leq \lim\limits_{\theta\to 0} \frac{\sin\theta}{\theta} \leq 1 \]

Clearly this means that \( \lim\limits_{\theta\to 0} \frac{\sin\theta}{\theta}=1\)

Two notes about the previous example are worth mentioning. First, one might be discouraged by this application, thinking "I would never have come up with that on my own. This is too hard!'' Don't be discouraged; within this text we will guide you in your use of the Squeeze Theorem. As one gains mathematical maturity, clever proofs like this are easier and easier to create.

Second, this limit tells us more than just that as \(x\) approaches 0, \(\sin(x)/x\) approaches 1. Both \(x\) and \(\sin x\) are approaching 0, but the ratio of \(x\) and \(\sin x\) approaches 1, meaning that they are approaching 0 in essentially the same way. Another way of viewing this is: for small \(x\), the functions \(y=x\) and \(y=\sin x\) are essentially indistinguishable.

We include this special limit, along with three others, in the following theorem.

Theorem \(\PageIndex{5}\) : Special Limits

• \( \lim\limits_{x\to 0} \frac{\sin x}{x} = 1\)
• \( \lim\limits_{x\to 0} \frac{\cos x-1}{x} = 0\)
• \( \lim\limits_{x\to 0} (1+x)^\frac1x = e\)
• \( \lim\limits_{x\to 0} \frac{e^x-1}{x} = 1\)

A short word on how to interpret the latter three limits. We know that as \(x\) goes to 0, \(\cos x\) goes to 1. So, in the second limit, both the numerator and denominator are approaching 0. However, since the limit is 0, we can interpret this as saying that "\(\cos x\) is approaching 1 faster than \(x\) is approaching 0.''

In the third limit, inside the parentheses we have an expression that is approaching 1 (though never equaling 1), and we know that 1 raised to any power is still 1. At the same time, the power is growing toward infinity. What happens to a number near 1 raised to a very large power? In this particular case, the result approaches Euler's number, \(e\), approximately \(2.718.\)

In the fourth limit, we see that as \(x\to 0\), \(e^x\) approaches 1 ``just as fast'' as \(x\to 0\), resulting in a limit of 1.

Our final theorem for this section will be motivated by the following example.

Example \(\PageIndex{5}\): Using algebra to evaluate a limit

Evaluate the following limit:

\[\lim\limits_{x\to 1}\frac{x^2-1}{x-1}. \nonumber\]

We begin by attempting to apply Theorem 3 and substituting 1 for \(x\) in the quotient. This gives:

\[\lim\limits_{x\to 1}\frac{x^2-1}{x-1} = \frac{1^2-1}{1-1} = \frac{"\,0\,"}{0},\]

and indeterminate form. We cannot apply the theorem.

By graphing the function, as in Figure \(\PageIndex{2}\), we see that the function seems to be linear, implying that the limit should be easy to evaluate. Recognize that the numerator of our quotient can be factored:

\[\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}. \nonumber\]

The function is not defined when \(x=1\), but for all other \(x\),

\[\require{cancel} \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = \frac{\cancel{(x-1)}(x+1)}{\cancel{x-1}}= x+1.\]

Clearly \( \lim\limits_{x\to 1}x+1 = 2\). Recall that when considering limits, we are not concerned with the value of the function at 1, only the value the function approaches as \(x\) approaches 1. Since \((x^2-1)/(x-1)\) and \(x+1\) are the same at all points except \(x=1\), they both approach the same value as \(x\) approaches 1. Therefore we can conclude that

\[\lim\limits_{x\to 1}\frac{x^2-1}{x-1}=2. \nonumber\]

The key to the above example is that the functions \(y=(x^2-1)/(x-1)\) and \(y=x+1\) are identical except at \(x=1\). Since limits describe a value the function is approaching, not the value the function actually attains, the limits of the two functions are always equal.

Theorem \(\PageIndex{6}\) : Limits of Functions Equal At All But One Point

Let \(g(x) = f(x)\) for all \(x\) in an open interval, except possibly at \(c\), and let \( \lim\limits_{x\to c} g(x) = L\) for some real number \(L\).Then

\[\lim\limits_{x\to c}f(x) = L.\]

The Fundamental Theorem of Algebra tells us that when dealing with a rational function of the form \(g(x)/f(x)\) and directly evaluating the limit \( \lim\limits_{x\to c} \frac{g(x)}{f(x)}\) returns "0/0'', then \((x-c)\) is a factor of both \(g(x)\) and \(f(x)\). One can then use algebra to factor this term out, cancel, then apply Theorem 6. We demonstrate this once more.

Example \(\PageIndex{6}\): Evaluating a limit using Theorem \(\PageIndex{6}\)

\[ \lim\limits_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15}. \nonumber\]

We begin by applying Theorem 3 and substituting 3 for \(x\).This returns the familiar indeterminate form of "0/0''.

Since the numerator and denominator are each polynomials, we know that \((x-3)\) is factor of each. Using whatever method is most comfortable to you, factor out \((x-3)\) from each (using polynomial division, synthetic division, a computer algebra system, etc.). We find that

\[\frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} = \frac{(x-3)(x^2+x-2)}{(x-3)(2 x^2+9 x-5)}. \nonumber\]

We can cancel the \((x-3)\) terms as long as \(x\neq 3\). Using Theorem 6 we conclude:

\[\begin{align*}\lim\limits_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} &= \lim\limits_{x\to 3}\frac{(x-3)(x^2+x-2)}{(x-3)(2 x^2+9 x-5)} \\ &= \lim\limits_{x\to 3} \frac{(x^2+x-2)}{(2 x^2+9 x-5)}\\ &= \frac{10}{40} = \frac14. \end{align*}\]

We end this section by revisiting a limit first seen in Section 1.1, a limit of a difference quotient. Let \(f(x) = -1.5x^2+11.5x\); we approximated the limit \( \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}\approx 8.5.\) We formally evaluate this limit in the following example.

Example \(\PageIndex{7}\) : Evaluating the limit of a difference quotient

Let \(f(x) = -1.5x^2+11.5x\); find \( \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}.\)

Since \(f\) is a polynomial, our first attempt should be to employ Theorem 3 and substitute 0 for \(h\). However, we see that this gives us "0/0.'' Knowing that we have a rational function hints that some algebra will help. Consider the following steps:

\[\begin{align*} \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h} &= \lim\limits_{h\to 0}\frac{-1.5(1+h)^2 + 11.5(1+h) - \left(-1.5(1)^2+11.5(1)\right)}{h} \\ &= \lim\limits_{h\to 0}\frac{-1.5(1+2h+h^2) + 11.5+11.5h - 10}{h}\\ &= \lim\limits_{h\to 0}\frac{-1.5h^2 +8.5h}{h}\\ &= \lim\limits_{h\to 0}\frac{h(-1.5h+8.5)}h\\ &= \lim\limits_{h\to 0}(-1.5h+8.5) \quad (\text{using Theorem 6, as \(h\neq 0\)}) \\ &= 8.5 \quad (\text{using Theorem 3}) \end{align*}\]

This matches our previous approximation.

This section contains several valuable tools for evaluating limits. One of the main results of this section is Theorem 3; it states that many functions that we use regularly behave in a very nice, predictable way. In the next section we give a name to this nice behavior; we label such functions as continuous . Defining that term will require us to look again at what a limit is and what causes limits to not exist.

Contributors and Attributions

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License.  http://www.apexcalculus.com/

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Limits by Rationalization

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We have seen several methods for finding limits , including limits by substitution , limits by factoring , and using the epsilon-delta definition of the limit .

In the case when direct substitution into the function gives an indeterminate form \(\big(\)such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\big)\) and the function involves a radical expression or a trigonometric function, it may be possible to find the limit by multiplying by a conjugate.

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Why RAG won’t solve generative AI’s hallucination problem

Hallucinations — the lies generative AI models tell, basically — are a big problem for businesses looking to integrate the technology into their operations.

Because models have no real intelligence and are simply predicting words, images, speech, music and other data according to a private schema , they sometimes get it wrong. Very wrong. In a recent piece in The Wall Street Journal, a source recounts an instance where Microsoft’s generative AI invented meeting attendees and implied that conference calls were about subjects that weren’t actually discussed on the call.

As I wrote a while ago, hallucinations may be an unsolvable problem with today’s transformer-based model architectures. But a number of generative AI vendors suggest that they can be done away with, more or less, through a technical approach called retrieval augmented generation, or RAG.

Here’s how one vendor, Squirro, pitches it :

At the core of the offering is the concept of Retrieval Augmented LLMs or Retrieval Augmented Generation (RAG) embedded in the solution … [our generative AI] is unique in its promise of zero hallucinations. Every piece of information it generates is traceable to a source, ensuring credibility.

Here’s a similar pitch from SiftHub:

Using RAG technology and fine-tuned large language models with industry-specific knowledge training, SiftHub allows companies to generate personalized responses with zero hallucinations. This guarantees increased transparency and reduced risk and inspires absolute trust to use AI for all their needs.

RAG was pioneered by data scientist Patrick Lewis, researcher at Meta and University College London, and lead author of the 2020 paper that coined the term. Applied to a model, RAG retrieves documents possibly relevant to a question — for example, a Wikipedia page about the Super Bowl — using what’s essentially a keyword search and then asks the model to generate answers given this additional context.

“When you’re interacting with a generative AI model like ChatGPT or Llama and you ask a question, the default is for the model to answer from its ‘parametric memory’ — i.e., from the knowledge that’s stored in its parameters as a result of training on massive data from the web,” David Wadden, a research scientist at AI2, the AI-focused research division of the nonprofit Allen Institute, explained. “But, just like you’re likely to give more accurate answers if you have a reference [like a book or a file] in front of you, the same is true in some cases for models.”

RAG is undeniably useful — it allows one to attribute things a model generates to retrieved documents to verify their factuality (and, as an added benefit, avoid potentially copyright-infringing regurgitation ). RAG also lets enterprises that don’t want their documents used to train a model — say, companies in highly regulated industries like healthcare and law — to allow models to draw on those documents in a more secure and temporary way.

But RAG certainly  can’t stop a model from hallucinating. And it has limitations that many vendors gloss over.

Wadden says that RAG is most effective in “knowledge-intensive” scenarios where a user wants to use a model to address an “information need” — for example, to find out who won the Super Bowl last year. In these scenarios, the document that answers the question is likely to contain many of the same keywords as the question (e.g., “Super Bowl,” “last year”), making it relatively easy to find via keyword search.

Things get trickier with “reasoning-intensive” tasks such as coding and math, where it’s harder to specify in a keyword-based search query the concepts needed to answer a request — much less identify which documents might be relevant.

Even with basic questions, models can get “distracted” by irrelevant content in documents, particularly in long documents where the answer isn’t obvious. Or they can — for reasons as yet unknown — simply ignore the contents of retrieved documents, opting instead to rely on their parametric memory.

RAG is also expensive in terms of the hardware needed to apply it at scale.

That’s because retrieved documents, whether from the web, an internal database or somewhere else, have to be stored in memory — at least temporarily — so that the model can refer back to them. Another expenditure is compute for the increased context a model has to process before generating its response. For a technology already notorious for the amount of compute and electricity it requires even for basic operations, this amounts to a serious consideration.

That’s not to suggest RAG can’t be improved. Wadden noted many ongoing efforts to train models to make better use of RAG-retrieved documents.

Some of these efforts involve models that can “decide” when to make use of the documents, or models that can choose not to perform retrieval in the first place if they deem it unnecessary. Others focus on ways to more efficiently index massive datasets of documents, and on improving search through better representations of documents — representations that go beyond keywords.

“We’re pretty good at retrieving documents based on keywords, but not so good at retrieving documents based on more abstract concepts, like a proof technique needed to solve a math problem,” Wadden said. “ Research is needed to build document representations and search techniques that can identify relevant documents for more abstract generation tasks. I think this is mostly an open question at this point.”

So RAG can help reduce a model’s hallucinations — but it’s not the answer to all of AI’s hallucinatory problems. Beware of any vendor that tries to claim otherwise.

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The limit does indeed exist in the Mean Girls remake, thus proving its inferiority to the original

20 years ago today, "Mean Girls" premiered and math problems were never the same again.

Lester Fabian Brathwaite is a staff writer at Entertainment Weekly , where he covers breaking news, all things Real Housewives , and a rich cornucopia of popular culture. Formerly a senior editor at Out magazine, his work has appeared on NewNowNext , Queerty , Rolling Stone , and The New Yorker . He was also the first author signed to Phoebe Robinson's Tiny Reparations imprint. He met Oprah once.

This year's Mean Girls remake did okay. Reviews were okay , the box office was okay . The general feeling was, "It's okay." But when competing with a film that looms so large in the cultural imagination, there's bound to be some disappointment with a new, arguably fresh take.

However, one glaring (for the mathematically-inclined) error in the remake offers proof, sorta, that it is inferior to the 2004 film, which celebrates its 20th anniversary today.

Paramount/ Everett; Jojo Whilden/Paramount

For fans of the Mean Girls universe, "The limit does not exist" is one of many iconic lines from the OG film. Spoken by Lindsay Lohan 's Cady Heron, the quote refers to a math problem she and her fellow Mathletes are tasked with solving at the big climactic state finals. In the scene, Cady accepts that she's become that which she hated and feared most, a mean girl, and begins her redemptoion arc.

In the 2024 film, Angourie Rice 's Cady Heron has the same realization, but the math problem she answers is slightly different than the one Lohan's Cady had. The limit does exist.

This video by YouTuber Oblivious Be n goes into depth about why and how the remake gets the problem wrong, but the basic explanation is that there's a plus sign after the zero (below), thus changing the equation altogether.

While not a life-changing error, Oblivious Ben argues that this is actually endemic with what's wrong with the remake — and by extension, reboots and remakes in general — in the first place.

"This juxtaposition shows just how much care went into the 2004 version of the movie versus how little care went into the 2024 version of the movie," he explains. "This new Mean Girls is more concerned with regurgitating lines and references from the original that it forgets what makes the original so special."

Citing the remake's problem as one of "reiteration," he continues, "The original version had been iterated and reiterated on so many times, that it's more important to reference the original than to focus on the version you're currently making."

And this is how math got to the core of what's wrong with Hollywood today. The limit does exist and that limit is apparently the third generation adaptation of a beloved pre-existing intellectual property.

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Related content:

• Lindsay Lohan pushes the 'limit' of iconic Mean Girls quote in new ad
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• Tina Fey on whether both Mean Girls movies share a universe

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