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L’Hôpital’s Rule

4.8 L’Hôpital’s Rule

Learning Objectives

  • 4.8.1 Recognize when to apply L’Hôpital’s rule.
  • 4.8.2 Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case.
  • 4.8.3 Describe the relative growth rates of functions.

In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule, uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value.

Applying L’Hôpital’s Rule

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

limx→af(x)g(x).limx→af(x)g(x).

If limx→af(x)=L1andlimx→ag(x)=L2≠0,limx→af(x)=L1andlimx→ag(x)=L2≠0, then

limx→af(x)g(x)=L1L2.limx→af(x)g(x)=L1L2.

However, what happens if limx→af(x)=0limx→af(x)=0 and limx→ag(x)=0?limx→ag(x)=0? We call this one of the indeterminate forms, of type 00.00. This is considered an indeterminate form because we cannot determine the exact behavior of f(x)g(x)f(x)g(x) as x→ax→a without further analysis. We have seen examples of this earlier in the text. For example, consider

limx→2x2−4x−2andlimx→0sinxx.limx→2x2−4x−2andlimx→0sinxx.

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

limx→2x2−4x−2=limx→2(x+2)(x−2)x−2=limx→2(x+2)=2+2=4.limx→2x2−4x−2=limx→2(x+2)(x−2)x−2=limx→2(x+2)=2+2=4.

For limx→0sinxxlimx→0sinxx we were able to show, using a geometric argument, that

limx→0sinxx=1.limx→0sinxx=1.

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions ff and gg such that limx→af(x)=0=limx→ag(x)limx→af(x)=0=limx→ag(x) and such that g′(a)≠0g′(a)≠0 For xx near a,a, we can write

f(x)≈f(a)+f′(a)(x−a)f(x)≈f(a)+f′(a)(x−a)

and

g(x)≈g(a)+g′(a)(x−a).g(x)≈g(a)+g′(a)(x−a).

Therefore,

f(x)g(x)≈f(a)+f′(a)(x−a)g(a)+g′(a)(x−a).f(x)g(x)≈f(a)+f′(a)(x−a)g(a)+g′(a)(x−a).
Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f’(a)(x – a) and y = g(a) + g’(a)(x – a) are also drawn.
Figure 4.71 If limx→af(x)=limx→ag(x),limx→af(x)=limx→ag(x), then the ratio f(x)/g(x)f(x)/g(x) is approximately equal to the ratio of their linear approximations near a.a.

Since ff is differentiable at a,a, then ff is continuous at a,a, and therefore f(a)=limx→af(x)=0.f(a)=limx→af(x)=0. Similarly, g(a)=limx→ag(x)=0.g(a)=limx→ag(x)=0. If we also assume that f′f′ and g′g′ are continuous at x=a,x=a, then f′(a)=limx→af′(x)f′(a)=limx→af′(x) and g′(a)=limx→ag′(x).g′(a)=limx→ag′(x). Using these ideas, we conclude that

limx→af(x)g(x)=limx→af′(x)(x−a)g′(x)(x−a)=limx→af′(x)g′(x).limx→af(x)g(x)=limx→af′(x)(x−a)g′(x)(x−a)=limx→af′(x)g′(x).

Note that the assumption that f′f′ and g′g′ are continuous at aa and g′(a)≠0g′(a)≠0 can be loosened. We state L’Hôpital’s rule formally for the indeterminate form 00.00. Also note that the notation 0000 does not mean we are actually dividing zero by zero. Rather, we are using the notation 0000 to represent a quotient of limits, each of which is zero.

Theorem 4.12

L’Hôpital’s Rule (0/0 Case)

Suppose ff and gg are differentiable functions over an open interval containing a,a, except possibly at a.a. If limx→af(x)=0limx→af(x)=0 and limx→ag(x)=0,limx→ag(x)=0, then

limx→af(x)g(x)=limx→af′(x)g′(x),limx→af(x)g(x)=limx→af′(x)g′(x),

assuming the limit on the right exists or is ∞∞ or −∞.−∞. This result also holds if we are considering one-sided limits, or if a=∞and−∞.a=∞and−∞.

Proof

We provide a proof of this theorem in the special case when f,g,f′,f,g,f′, and g′g′ are all continuous over an open interval containing a.a. In that case, since limx→af(x)=0=limx→ag(x)limx→af(x)=0=limx→ag(x) and ff and gg are continuous at a,a, it follows that f(a)=0=g(a).f(a)=0=g(a). Therefore,

limx→af(x)g(x)=limx→af(x)−f(a)g(x)−g(a)sincef(a)=0=g(a)=limx→af(x)−f(a)x−ag(x)−g(a)x−aalgebra=limx→af(x)−f(a)x−alimx→ag(x)−g(a)x−alimit of a quotient=f′(a)g′(a)definition of the derivative=limx→af′(x)limx→ag′(x)continuity off′andg′=limx→af′(x)g′(x).limit of a quotientlimx→af(x)g(x)=limx→af(x)−f(a)g(x)−g(a)sincef(a)=0=g(a)=limx→af(x)−f(a)x−ag(x)−g(a)x−aalgebra=limx→af(x)−f(a)x−alimx→ag(x)−g(a)x−alimit of a quotient=f′(a)g′(a)definition of the derivative=limx→af′(x)limx→ag′(x)continuity off′andg′=limx→af′(x)g′(x).limit of a quotient

Note that L’Hôpital’s rule states we can calculate the limit of a quotient fgfg by considering the limit of the quotient of the derivatives f′g′.f′g′. It is important to realize that we are not calculating the derivative of the quotient fg.fg.

□

Example 4.38

Applying L’Hôpital’s Rule (0/0 Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

  1. limx→01−cosxxlimx→01−cosxx
  2. limx→1sin(πx)lnxlimx→1sin(πx)lnx
  3. limx→∞e1/x−11/xlimx→∞e1/x−11/x
  4. limx→0sinx−xx2limx→0sinx−xx2

Solution

  1. Since the numerator 1−cosx→01−cosx→0 and the denominator x→0,x→0, we can apply L’Hôpital’s rule to evaluate this limit. We have
    limx→01−cosxx=limx→0ddx(1−cosx)ddx(x)=limx→0sinx1=limx→0(sinx)limx→0(1)=01=0.limx→01−cosxx=limx→0ddx(1−cosx)ddx(x)=limx→0sinx1=limx→0(sinx)limx→0(1)=01=0.
  2. As x→1,x→1, the numerator sin(πx)→0sin(πx)→0 and the denominator ln(x)→0.ln(x)→0. Therefore, we can apply L’Hôpital’s rule. We obtain
    limx→1sin(πx)lnx=limx→1πcos(πx)1/x=limx→1(πx)cos(πx)=(π·1)(−1)=−π.limx→1sin(πx)lnx=limx→1πcos(πx)1/x=limx→1(πx)cos(πx)=(π·1)(−1)=−π.
  3. As x→∞,x→∞, the numerator e1/x−1→0e1/x−1→0 and the denominator (1x)→0.(1x)→0. Therefore, we can apply L’Hôpital’s rule. We obtain
    limx→∞e1/x−11x=limx→∞e1/x(−1x2)(−1x2)=limx→∞e1/x=e0=1.limx→∞e1/x−11x=limx→∞e1/x(−1x2)(−1x2)=limx→∞e1/x=e0=1.
  4. As x→0,x→0, both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
    limx→0sinx−xx2=limx→0cosx−12x.limx→0sinx−xx2=limx→0cosx−12x.

     Since the numerator and denominator of this new quotient both approach zero as x→0,x→0, we apply L’Hôpital’s rule again. In doing so, we see that
    limx→0cosx−12x=limx→0−sinx2=0.limx→0cosx−12x=limx→0−sinx2=0.

     Therefore, we conclude that
    limx→0sinx−xx2=0.limx→0sinx−xx2=0.

Checkpoint 4.37

Evaluate limx→0xtanx.limx→0xtanx.

We can also use L’Hôpital’s rule to evaluate limits of quotients f(x)g(x)f(x)g(x) in which f(x)→±∞f(x)→±∞ and g(x)→±∞.g(x)→±∞. Limits of this form are classified as indeterminate forms of type ∞/∞.∞/∞. Again, note that we are not actually dividing ∞∞ by ∞.∞. Since ∞∞ is not a real number, that is impossible; rather, ∞/∞.∞/∞. is used to represent a quotient of limits, each of which is ∞∞ or −∞.−∞.

Theorem 4.13

L’Hôpital’s Rule (∞/∞(∞/∞ Case)

Suppose ff and gg are differentiable functions over an open interval containing a,a, except possibly at a.a. Suppose limx→af(x)=∞limx→af(x)=∞ (or −∞)−∞) and limx→ag(x)=∞limx→ag(x)=∞ (or −∞).−∞). Then,

limx→af(x)g(x)=limx→af′(x)g′(x),limx→af(x)g(x)=limx→af′(x)g′(x),

assuming the limit on the right exists or is ∞∞ or −∞.−∞. This result also holds if the limit is infinite, if a=∞a=∞ or −∞,−∞, or the limit is one-sided.

Example 4.39

Applying L’Hôpital’s Rule (∞/∞(∞/∞ Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

  1. limx→∞3x+52x+1limx→∞3x+52x+1
  2. limx→0+lnxcotxlimx→0+lnxcotx

Solution

  1. Since 3x+53x+5 and 2x+12x+1 are first-degree polynomials with positive leading coefficients, limx→∞(3x+5)=∞limx→∞(3x+5)=∞ and limx→∞(2x+1)=∞.limx→∞(2x+1)=∞. Therefore, we apply L’Hôpital’s rule and obtain
    limx→∞3x+52x+1=limx→∞32=32.limx→∞3x+52x+1=limx→∞32=32.

     Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of xx in the denominator. In doing so, we saw that
    limx→∞3x+52x+1=limx→∞3+5/x2+1/x=32.limx→∞3x+52x+1=limx→∞3+5/x2+1/x=32.

     L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
  2. Here, limx→0+lnx=−∞limx→0+lnx=−∞ and limx→0+cotx=∞.limx→0+cotx=∞. Therefore, we can apply L’Hôpital’s rule and obtain
    limx→0+lnxcotx=limx→0+1/x−csc2x=limx→0+1−xcsc2x.limx→0+lnxcotx=limx→0+1/x−csc2x=limx→0+1−xcsc2x.

     Now as x→0+,x→0+, csc2x→∞.csc2x→∞. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of cscxcscx to write
    limx→0+1−xcsc2x=limx→0+sin2x−x.limx→0+1−xcsc2x=limx→0+sin2x−x.

     Now limx→0+sin2x=0limx→0+sin2x=0 and limx→0+x=0,limx→0+x=0, so we apply L’Hôpital’s rule again. We find
    limx→0+sin2x−x=limx→0+2sinxcosx−1=0−1=0.limx→0+sin2x−x=limx→0+2sinxcosx−1=0−1=0.

     We conclude that
    limx→0+lnxcotx=0.limx→0+lnxcotx=0.

Checkpoint 4.38

Evaluate limx→∞lnx5x.limx→∞lnx5x.

As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient f(x)g(x),f(x)g(x), it is essential that the limit of f(x)g(x)f(x)g(x) be of the form 0000 or ∞/∞.∞/∞. Consider the following example.

Example 4.40

When L’Hôpital’s Rule Does Not Apply

Consider limx→1x2+53x+4.limx→1x2+53x+4. Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

Solution

Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get

d d x ( x 2 + 5 ) = 2 x d d x ( x 2 + 5 ) = 2 x

and

d d x ( 3 x + 4 ) = 3 . d d x ( 3 x + 4 ) = 3 .

At which point we would conclude erroneously that

lim x → 1 x 2 + 5 3 x + 4 = lim x → 1 2 x 3 = 2 3 . lim x → 1 x 2 + 5 3 x + 4 = lim x → 1 2 x 3 = 2 3 .

However, since limx→1(x2+5)=6limx→1(x2+5)=6 and limx→1(3x+4)=7,limx→1(3x+4)=7, we actually have

lim x → 1 x 2 + 5 3 x + 4 = 6 7 . lim x → 1 x 2 + 5 3 x + 4 = 6 7 .

We can conclude that

lim x → 1 x 2 + 5 3 x + 4 ≠ lim x → 1 d d x ( x 2 + 5 ) d d x ( 3 x + 4 ) . lim x → 1 x 2 + 5 3 x + 4 ≠ lim x → 1 d d x ( x 2 + 5 ) d d x ( 3 x + 4 ) .

Checkpoint 4.39

Explain why we cannot apply L’Hôpital’s rule to evaluate limx→0+cosxx.limx→0+cosxx. Evaluate limx→0+cosxxlimx→0+cosxx by other means.

Other Indeterminate Forms

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms 0000 and ∞/∞.∞/∞. However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions 0·∞,0·∞, ∞−∞,∞−∞, 1∞,1∞, ∞0,∞0, and 0000 are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form 0000 or ∞/∞.∞/∞.

Indeterminate Form of Type 0·∞0·∞

Suppose we want to evaluate limx→a(f(x)·g(x)),limx→a(f(x)·g(x)), where f(x)→0f(x)→0 and g(x)→∞g(x)→∞ (or −∞)−∞) as x→a.x→a. Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation 0·∞0·∞ to denote the form that arises in this situation. The expression 0·∞0·∞ is considered indeterminate because we cannot determine without further analysis the exact behavior of the product f(x)g(x)f(x)g(x) as x→a.x→a. For example, let nn be a positive integer and consider

f(x)=1(xn+1)andg(x)=3x2.f(x)=1(xn+1)andg(x)=3x2.

As x→∞,x→∞, f(x)→0f(x)→0 and g(x)→∞.g(x)→∞. However, the limit as x→∞x→∞ of f(x)g(x)=3x2(xn+1)f(x)g(x)=3x2(xn+1) varies, depending on n.n. If n=2,n=2, then limx→∞f(x)g(x)=3.limx→∞f(x)g(x)=3. If n=1,n=1, then limx→∞f(x)g(x)=∞.limx→∞f(x)g(x)=∞. If n=3,n=3, then limx→∞f(x)g(x)=0.limx→∞f(x)g(x)=0. Here we consider another limit involving the indeterminate form 0·∞0·∞ and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

Example 4.41

Indeterminate Form of Type 0·∞0·∞

Evaluate limx→0+xlnx.limx→0+xlnx.

Solution

First, rewrite the function xlnxxlnx as a quotient to apply L’Hôpital’s rule. If we write

x ln x = ln x 1 / x , x ln x = ln x 1 / x ,

we see that lnx→−∞lnx→−∞ as x→0+x→0+ and 1x→∞1x→∞ as x→0+.x→0+. Therefore, we can apply L’Hôpital’s rule and obtain

lim x → 0 + ln x 1 / x = lim x → 0 + d d x ( ln x ) d d x ( 1 / x ) = lim x → 0 + 1 / x -1 / x 2 = lim x → 0 + ( − x ) = 0 . lim x → 0 + ln x 1 / x = lim x → 0 + d d x ( ln x ) d d x ( 1 / x ) = lim x → 0 + 1 / x -1 / x 2 = lim x → 0 + ( − x ) = 0 .

We conclude that

lim x → 0 + x ln x = 0 . lim x → 0 + x ln x = 0 .
The function y = x ln(x) is graphed for values x ≥ 0. At x = 0, the value of the function is 0.
Figure 4.72 Finding the limit at x=0x=0 of the function f(x)=xlnx.f(x)=xlnx.

Checkpoint 4.40

Evaluate limx→0xcotx.limx→0xcotx.

Indeterminate Form of Type ∞−∞∞−∞

Another type of indeterminate form is ∞−∞.∞−∞. Consider the following example. Let nn be a positive integer and let f(x)=3xnf(x)=3xn and g(x)=3x2+5.g(x)=3x2+5. As x→∞,x→∞, f(x)→∞f(x)→∞ and g(x)→∞.g(x)→∞. We are interested in limx→∞(f(x)−g(x)).limx→∞(f(x)−g(x)). Depending on whether f(x)f(x) grows faster, g(x)g(x) grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since f(x)→∞f(x)→∞ and g(x)→∞,g(x)→∞, we write ∞−∞∞−∞ to denote the form of this limit. As with our other indeterminate forms, ∞−∞∞−∞ has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent nn in the function f(x)=3xnf(x)=3xn is n=3,n=3, then

limx→∞(f(x)−g(x))=limx→∞(3x3−3x2−5)=∞.limx→∞(f(x)−g(x))=limx→∞(3x3−3x2−5)=∞.

On the other hand, if n=2,n=2, then

limx→∞(f(x)−g(x))=limx→∞(3x2−3x2−5)=−5.limx→∞(f(x)−g(x))=limx→∞(3x2−3x2−5)=−5.

However, if n=1,n=1, then

limx→∞(f(x)−g(x))=limx→∞(3x−3x2−5)=−∞.limx→∞(f(x)−g(x))=limx→∞(3x−3x2−5)=−∞.

Therefore, the limit cannot be determined by considering only ∞−∞.∞−∞. Next we see how to rewrite an expression involving the indeterminate form ∞−∞∞−∞ as a fraction to apply L’Hôpital’s rule.

Example 4.42

Indeterminate Form of Type ∞−∞∞−∞

Evaluate limx→0+(1x2−1tanx).limx→0+(1x2−1tanx).

Solution

By combining the fractions, we can write the function as a quotient. Since the least common denominator is x2tanx,x2tanx, we have

1 x 2 − 1 tan x = ( tan x ) − x 2 x 2 tan x . 1 x 2 − 1 tan x = ( tan x ) − x 2 x 2 tan x .

As x→0+,x→0+, the numerator tanx−x2→0tanx−x2→0 and the denominator x2tanx→0.x2tanx→0. Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

lim x → 0 + ( tan x ) − x 2 x 2 tan x = lim x → 0 + ( sec 2 x ) − 2 x x 2 sec 2 x + 2 x tan x . lim x → 0 + ( tan x ) − x 2 x 2 tan x = lim x → 0 + ( sec 2 x ) − 2 x x 2 sec 2 x + 2 x tan x .

As x→0+,x→0+, (sec2x)−2x→1(sec2x)−2x→1 and x2sec2x+2xtanx→0.x2sec2x+2xtanx→0. Since the denominator is positive as xx approaches zero from the right, we conclude that

lim x → 0 + ( sec 2 x ) − 2 x x 2 sec 2 x + 2 x tan x = ∞ . lim x → 0 + ( sec 2 x ) − 2 x x 2 sec 2 x + 2 x tan x = ∞ .

Therefore,

lim x → 0 + ( 1 x 2 − 1 tan x ) = ∞ . lim x → 0 + ( 1 x 2 − 1 tan x ) = ∞ .

Checkpoint 4.41

Evaluate limx→0+(1x−1sinx).limx→0+(1x−1sinx).

Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions 00,00, ∞0,∞0, and 1∞1∞ are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms.

Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate limx→af(x)g(x)limx→af(x)g(x) and we arrive at the indeterminate form ∞0.∞0. (The indeterminate forms 0000 and 1∞1∞ can be handled similarly.) We proceed as follows. Let

y=f(x)g(x).y=f(x)g(x).

Then,

lny=ln(f(x)g(x))=g(x)ln(f(x)).lny=ln(f(x)g(x))=g(x)ln(f(x)).

Therefore,

limx→a[ln(y)]=limx→a[g(x)ln(f(x))].limx→a[ln(y)]=limx→a[g(x)ln(f(x))].

Since limx→af(x)=∞,limx→af(x)=∞, we know that limx→aln(f(x))=∞.limx→aln(f(x))=∞. Therefore, limx→ag(x)ln(f(x))limx→ag(x)ln(f(x)) is of the indeterminate form 0·∞,0·∞, and we can use the techniques discussed earlier to rewrite the expression g(x)ln(f(x))g(x)ln(f(x)) in a form so that we can apply L’Hôpital’s rule. Suppose limx→ag(x)ln(f(x))=L,limx→ag(x)ln(f(x))=L, where LL may be ∞∞ or −∞.−∞. Then

limx→a[ln(y)]=L.limx→a[ln(y)]=L.

Since the natural logarithm function is continuous, we conclude that

ln(limx→ay)=L,ln(limx→ay)=L,

which gives us

limx→ay=limx→af(x)g(x)=eL.limx→ay=limx→af(x)g(x)=eL.

Example 4.43

Indeterminate Form of Type ∞0∞0

Evaluate limx→∞x1/x.limx→∞x1/x.

Solution

Let y=x1/x.y=x1/x. Then,

ln y = ln ( x 1 / x ) = 1 x ln x = ln x x . ln y = ln ( x 1 / x ) = 1 x ln x = ln x x .

We need to evaluate limx→∞lnxx.limx→∞lnxx. Applying L’Hôpital’s rule, we obtain

lim x → ∞ ln y = lim x → ∞ ln x x = lim x → ∞ 1 / x 1 = 0 . lim x → ∞ ln y = lim x → ∞ ln x x = lim x → ∞ 1 / x 1 = 0 .

Therefore, limx→∞lny=0.limx→∞lny=0. Since the natural logarithm function is continuous, we conclude that

ln ( lim x → ∞ y ) = 0 , ln ( lim x → ∞ y ) = 0 ,

which leads to

lim x → ∞ y = e ln ( y ) = lim   e 0 = 1 . (All limits   x   ≥   ∞ .) lim x → ∞ y = e ln ( y ) = lim   e 0 = 1 . (All limits   x   ≥   ∞ .)

Hence,

lim x → ∞ x 1 / x = 1 . lim x → ∞ x 1 / x = 1 .

Checkpoint 4.42

Evaluate limx→∞x1/ln(x).limx→∞x1/ln(x).

Example 4.44

Indeterminate Form of Type 0000

Evaluate limx→0+xsinx.limx→0+xsinx.

Solution

Let

y = x sin x . y = x sin x .

Therefore,

ln y = ln ( x sin x ) = sin x ln x . ln y = ln ( x sin x ) = sin x ln x .

We now evaluate limx→0+sinxlnx.limx→0+sinxlnx. Since limx→0+sinx=0limx→0+sinx=0 and limx→0+lnx=−∞,limx→0+lnx=−∞, we have the indeterminate form 0·∞.0·∞. To apply L’Hôpital’s rule, we need to rewrite sinxlnxsinxlnx as a fraction. We could write

sin x ln x = sin x 1 / ln x sin x ln x = sin x 1 / ln x

or

sin x ln x = ln x 1 / sin x = ln x csc x . sin x ln x = ln x 1 / sin x = ln x csc x .

Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

lim x → 0 + sin x ln x = lim x → 0 + sin x 1 / ln x = lim x → 0 + cos x −1 / ( x ( ln x ) 2 ) = lim x → 0 + ( − x ( ln x ) 2 cos x ) . lim x → 0 + sin x ln x = lim x → 0 + sin x 1 / ln x = lim x → 0 + cos x −1 / ( x ( ln x ) 2 ) = lim x → 0 + ( − x ( ln x ) 2 cos x ) .

Unfortunately, we not only have another expression involving the indeterminate form 0·∞,0·∞, but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing

sin x ln x = ln x 1 / sin x = ln x csc x , sin x ln x = ln x 1 / sin x = ln x csc x ,

and applying L’Hôpital’s rule, we obtain

lim x → 0 + sin x ln x = lim x → 0 + ln x csc x = lim x → 0 + 1 / x − csc x cot x = lim x → 0 + −1 x csc x cot x . lim x → 0 + sin x ln x = lim x → 0 + ln x csc x = lim x → 0 + 1 / x − csc x cot x = lim x → 0 + −1 x csc x cot x .

Using the fact that cscx=1sinxcscx=1sinx and cotx=cosxsinx,cotx=cosxsinx, we can rewrite the expression on the right-hand side as

lim x → 0 + − sin 2 x x cos x = lim x → 0 + [ sin x x · ( − tan x ) ] = ( lim x → 0 + sin x x ) · ( lim x → 0 + ( − tan x ) ) = 1 · 0 = 0 . lim x → 0 + − sin 2 x x cos x = lim x → 0 + [ sin x x · ( − tan x ) ] = ( lim x → 0 + sin x x ) · ( lim x → 0 + ( − tan x ) ) = 1 · 0 = 0 .

We conclude that limx→0+lny=0.limx→0+lny=0. Therefore, ln(limx→0+y)=0ln(limx→0+y)=0 and we have

lim x → 0 + y = lim x → 0 + x sin x = e 0 = 1 . lim x → 0 + y = lim x → 0 + x sin x = e 0 = 1 .

Hence,

lim x → 0 + x sin x = 1 . lim x → 0 + x sin x = 1 .

Checkpoint 4.43

Evaluate limx→0+xx.limx→0+xx.

Growth Rates of Functions

Suppose the functions ff and gg both approach infinity as x→∞.x→∞. Although the values of both functions become arbitrarily large as the values of xx become sufficiently large, sometimes one function is growing more quickly than the other. For example, f(x)=x2f(x)=x2 and g(x)=x3g(x)=x3 both approach infinity as x→∞.x→∞. However, as shown in the following table, the values of x3x3 are growing much faster than the values of x2.x2.

xx 1010 100100 10001000 10,00010,000
f(x)=x2f(x)=x2 100100 10,00010,000 1,000,0001,000,000 100,000,000100,000,000
g(x)=x3g(x)=x3 10001000 1,000,0001,000,000 1,000,000,0001,000,000,000 1,000,000,000,0001,000,000,000,000
Table 4.7 Comparing the Growth Rates of x 2 x 2 and x 3 x 3

In fact,

limx→∞x3x2=limx→∞x=∞.or, equivalently,limx→∞x2x3=limx→∞1x=0.limx→∞x3x2=limx→∞x=∞.or, equivalently,limx→∞x2x3=limx→∞1x=0.

As a result, we say x3x3 is growing more rapidly than x2x2 as x→∞.x→∞. On the other hand, for f(x)=x2f(x)=x2 and g(x)=3x2+4x+1,g(x)=3x2+4x+1, although the values of g(x)g(x) are always greater than the values of f(x)f(x) for x>0,x>0, each value of g(x)g(x) is roughly three times the corresponding value of f(x)f(x) as x→∞,x→∞, as shown in the following table. In fact,

limx→∞x23x2+4x+1=13.limx→∞x23x2+4x+1=13.
xx 1010 100100 10001000 10,00010,000
f(x)=x2f(x)=x2 100100 10,00010,000 1,000,0001,000,000 100,000,000100,000,000
g(x)=3x2+4x+1g(x)=3x2+4x+1 341341 30,40130,401 3,004,0013,004,001 300,040,001300,040,001
Table 4.8 Comparing the Growth Rates of x 2 x 2 and 3 x 2 + 4 x + 1 3 x 2 + 4 x + 1

In this case, we say that x2x2 and 3x2+4x+13x2+4x+1 are growing at the same rate as x→∞.x→∞.

More generally, suppose ff and gg are two functions that approach infinity as x→∞.x→∞. We say gg grows more rapidly than ff as x→∞x→∞ if

limx→∞g(x)f(x)=∞;or, equivalently,limx→∞f(x)g(x)=0.limx→∞g(x)f(x)=∞;or, equivalently,limx→∞f(x)g(x)=0.

On the other hand, if there exists a constant M≠0M≠0 such that

limx→∞f(x)g(x)=M,limx→∞f(x)g(x)=M,

we say ff and gg grow at the same rate as x→∞.x→∞.

Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.

Example 4.45

Comparing the Growth Rates of ln(x),ln(x), x2,x2, and exex

For each of the following pairs of functions, use L’Hôpital’s rule to evaluate limx→∞(f(x)g(x)).limx→∞(f(x)g(x)).

  1. f(x)=x2andg(x)=exf(x)=x2andg(x)=ex
  2. f(x)=ln(x)andg(x)=x2f(x)=ln(x)andg(x)=x2

Solution

  1. Since limx→∞x2=∞limx→∞x2=∞ and limx→∞ex=∞,limx→∞ex=∞, we can use L’Hôpital’s rule to evaluate limx→∞[x2ex].limx→∞[x2ex]. We obtain
    limx→∞x2ex=limx→∞2xex.limx→∞x2ex=limx→∞2xex.

     Since limx→∞2x=∞limx→∞2x=∞ and limx→∞ex=∞,limx→∞ex=∞, we can apply L’Hôpital’s rule again. Since
    limx→∞2xex=limx→∞2ex=0,limx→∞2xex=limx→∞2ex=0,

     we conclude that
    limx→∞x2ex=0.limx→∞x2ex=0.

     Therefore, exex grows more rapidly than x2x2 as x→∞x→∞ (See Figure 4.73 and Table 4.9).
    The functions g(x) = ex and f(x) = x2 are graphed. It is obvious that g(x) increases much more quickly than f(x).
    Figure 4.73 An exponential function grows at a faster rate than a power function.
    xx 55 1010 1515 2020
    x2x2 2525 100100 225225 400400
    exex 148148 22,02622,026 3,269,0173,269,017 485,165,195485,165,195
    Table 4.9 Growth rates of a power function and an exponential function.
  2. Since limx→∞lnx=∞limx→∞lnx=∞ and limx→∞x2=∞,limx→∞x2=∞, we can use L’Hôpital’s rule to evaluate limx→∞lnxx2.limx→∞lnxx2. We obtain
    limx→∞lnxx2=limx→∞1/x2x=limx→∞12x2=0.limx→∞lnxx2=limx→∞1/x2x=limx→∞12x2=0.

     Thus, x2x2 grows more rapidly than lnxlnx as x→∞x→∞ (see Figure 4.74 and Table 4.10).
    The functions g(x) = x2 and f(x) = ln(x) are graphed. It is obvious that g(x) increases much more quickly than f(x).
    Figure 4.74 A power function grows at a faster rate than a logarithmic function.
    xx 1010 100100 10001000 10,00010,000
    ln(x)ln(x) 2.3032.303 4.6054.605 6.9086.908 9.2109.210
    x2x2 100100 10,00010,000 1,000,0001,000,000 100,000,000100,000,000
    Table 4.10 Growth rates of a power function and a logarithmic function

Checkpoint 4.44

Compare the growth rates of x100x100 and 2x.2x.

Using the same ideas as in Example 4.45a. it is not difficult to show that exex grows more rapidly than xpxp for any p>0.p>0. In Figure 4.75 and Table 4.11, we compare exex with x3x3 and x4x4 as x→∞.x→∞.

This figure has two figures marked a and b. In figure a, the functions y = ex and y = x3 are graphed. It is obvious that ex increases more quickly than x3. In figure b, the functions y = ex and y = x4 are graphed. It is obvious that ex increases much more quickly than x4, but the point at which that happens is further to the right than it was for x3.
Figure 4.75 The exponential function exex grows faster than xpxp for any p>0.p>0. (a) A comparison of exex with x3.x3. (b) A comparison of exex with x4.x4.
xx 55 1010 1515 2020
x3x3 125125 10001000 33753375 80008000
x4x4 625625 10,00010,000 50,62550,625 160,000160,000
exex 148148 22,02622,026 3,269,0173,269,017 485,165,195485,165,195
Table 4.11 An exponential function grows at a faster rate than any power function

Similarly, it is not difficult to show that xpxp grows more rapidly than lnxlnx for any p>0.p>0. In Figure 4.76 and Table 4.12, we compare lnxlnx with x3x3 and x.x.

This figure shows y = the square root of x, y = the cube root of x, and y = ln(x). It is apparent that y = ln(x) grows more slowly than either of these functions.
Figure 4.76 The function y=ln(x)y=ln(x) grows more slowly than xpxp for any p>0p>0 as x→∞.x→∞.
xx 1010 100100 10001000 10,00010,000
ln(x)ln(x) 2.3032.303 4.6054.605 6.9086.908 9.2109.210
x3x3 2.1542.154 4.6424.642 1010 21.54421.544
xx 3.1623.162 1010 31.62331.623 100100
Table 4.12 A logarithmic function grows at a slower rate than any root function

Section 4.8 Exercises

For the following exercises, evaluate the limit.

356.

Evaluate the limit limx→∞exx.limx→∞exx.

357.

Evaluate the limit limx→∞exxk.limx→∞exxk.

358.

Evaluate the limit limx→∞lnxxk.limx→∞lnxxk.

359.

Evaluate the limit limx→ax−ax2−a2,a≠0limx→ax−ax2−a2,a≠0.

360.

Evaluate the limit limx→ax−ax3−a3,a≠0limx→ax−ax3−a3,a≠0.

361.

Evaluate the limit limx→ax−axn−an,a≠0limx→ax−axn−an,a≠0.

For the following exercises, determine whether you can apply L’Hôpital’s rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L’Hôpital’s rule.

362.

lim x → 0 + x 2 ln x lim x → 0 + x 2 ln x

363.

lim x → ∞ x 1 / x lim x → ∞ x 1 / x

364.

lim x → 0 x 2 / x lim x → 0 x 2 / x

365.

lim x → 0 x 2 1 / x lim x → 0 x 2 1 / x

366.

lim x → ∞ e x x lim x → ∞ e x x

For the following exercises, evaluate the limits with either L’Hôpital’s rule or previously learned methods.

367.

lim x → 3 x 2 − 9 x − 3 lim x → 3 x 2 − 9 x − 3

368.

lim x → 3 x 2 − 9 x + 3 lim x → 3 x 2 − 9 x + 3

369.

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

370.

lim x → π / 2 cos x π 2 − x lim x → π / 2 cos x π 2 − x

371.

lim x → π x − π sin x lim x → π x − π sin x

372.

lim x → 1 x − 1 sin x lim x → 1 x − 1 sin x

373.

lim x → 0 ( 1 + x ) n − 1 x lim x → 0 ( 1 + x ) n − 1 x

374.

lim x → 0 ( 1 + x ) n − 1 − n x x 2 lim x → 0 ( 1 + x ) n − 1 − n x x 2

375.

lim x → 0 sin x − tan x x 3 lim x → 0 sin x − tan x x 3

376.

lim x → 0 1 + x − 1 − x x lim x → 0 1 + x − 1 − x x

377.

lim x → 0 e x − x − 1 x 2 lim x → 0 e x − x − 1 x 2

378.

lim x → 0 + tan x x lim x → 0 + tan x x

379.

lim x → 1 x − 1 ln x lim x → 1 x − 1 ln x

380.

lim x → 0 ( x + 1 ) 1 / x lim x → 0 ( x + 1 ) 1 / x

381.

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

382.

lim x → 0 + x 2 x lim x → 0 + x 2 x

383.

lim x → ∞ x sin ( 1 x ) lim x → ∞ x sin ( 1 x )

384.

lim x → 0 sin x − x x 2 lim x → 0 sin x − x x 2

385.

lim x → 0 + x ln ( x 4 ) lim x → 0 + x ln ( x 4 )

386.

lim x → ∞ ( x − e x ) lim x → ∞ ( x − e x )

387.

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

388.

lim x → 0 3 x − 2 x x lim x → 0 3 x − 2 x x

389.

lim x → 0 1 + 1 / x 1 − 1 / x lim x → 0 1 + 1 / x 1 − 1 / x

390.

lim x → π / 4 ( 1 − tan x ) cot x lim x → π / 4 ( 1 − tan x ) cot x

391.

lim x → ∞ x e 1 / x lim x → ∞ x e 1 / x

392.

lim x → 0 + x 1 / cos x lim x → 0 + x 1 / cos x

393.

lim x → 0 + x 1 / x lim x → 0 + x 1 / x

394.

lim x → 0 - ( 1 − 1 x ) x lim x → 0 - ( 1 − 1 x ) x

395.

lim x → ∞ ( 1 − 1 x ) x lim x → ∞ ( 1 − 1 x ) x

For the following exercises, use a calculator to graph the function and estimate the value of the limit, then use L’Hôpital’s rule to find the limit directly.

396.

[T] limx→0ex−1xlimx→0ex−1x

397.

[T] limx→0xsin(1x)limx→0xsin(1x)

398.

[T] limx→1x−11−cos(πx)limx→1x−11−cos(πx)

399.

[T] limx→1e(x−1)−1x−1limx→1e(x−1)−1x−1

400.

[T] limx→1(x−1)2lnxlimx→1(x−1)2lnx

401.

[T] limx→π1+cosxsinxlimx→π1+cosxsinx

402.

[T] limx→0(cscx−1x)limx→0(cscx−1x)

403.

[T] limx→0+tan(xx)limx→0+tan(xx)

404.

[T] limx→0+lnxsinxlimx→0+lnxsinx

405.

[T] limx→0ex−e−xxlimx→0ex−e−xx