L’Hôpital’s Rule · Calculus (2024)

  • Recognize when to apply L’Hôpital’s rule.
  • Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case.
  • 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

limxaf(x)g(x).

If limxaf(x)=L1andlimxag(x)=L20,

then

limxaf(x)g(x)=L1L2.

However, what happens if limxaf(x)=0

and limxag(x)=0?

We call this one of the indeterminate forms, of type 00.

This is considered an indeterminate form because we cannot determine the exact behavior of f(x)g(x)

as xa

without further analysis. We have seen examples of this earlier in the text. For example, consider

limx2x24x2andlimx0sinxx.

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

limx2x24x2=limx2(x+2)(x2)x2=limx2(x+2)=2+2=4.

For limx0sinxx

we were able to show, using a geometric argument, that

limx0sinxx=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 f

and g

such that limxaf(x)=0=limxag(x)

and such that g(a)0

For x

near a,

we can write

f(x)f(a)+f(a)(xa)

and

g(x)g(a)+g(a)(xa).

Therefore,

f(x)g(x)f(a)+f(a)(xa)g(a)+g(a)(xa).

L’Hôpital’s Rule · Calculus (1)

Since f

is differentiable at a,

then f

is continuous at a,

and therefore f(a)=limxaf(x)=0.

Similarly, g(a)=limxag(x)=0.

If we also assume that f

and g

are continuous at x=a,

then f(a)=limxaf(x)

and g(a)=limxag(x).

Using these ideas, we conclude that

limxaf(x)g(x)=limxaf(x)(xa)g(x)(xa)=limxaf(x)g(x).

Note that the assumption that f

and g

are continuous at a

and g(a)0

can be loosened. We state L’Hôpital’s rule formally for the indeterminate form 00.

Also note that the notation 00

does not mean we are actually dividing zero by zero. Rather, we are using the notation 00

to represent a quotient of limits, each of which is zero.

L’Hôpital’s Rule (0/0 Case)

Suppose f

and g

are differentiable functions over an open interval containing a,

except possibly at a.

If limxaf(x)=0

and limxag(x)=0,

then

limxaf(x)g(x)=limxaf(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.

Proof

We provide a proof of this theorem in the special case when f,g,f,

and g

are all continuous over an open interval containing a.

In that case, since limxaf(x)=0=limxag(x)

and f

and g

are continuous at a,

it follows that f(a)=0=g(a).

Therefore,

limxaf(x)g(x)=limxaf(x)f(a)g(x)g(a)sincef(a)=0=g(a)=limxaf(x)f(a)xag(x)g(a)xaalgebra=limxaf(x)f(a)xalimxag(x)g(a)xalimit of a quotient=f(a)g(a)definition of the derivative=limxaf(x)limxag(x)continuity offandg=limxaf(x)g(x).limit of a quotient

Note that L’Hôpital’s rule states we can calculate the limit of a quotient fg

by considering the limit of the quotient of the derivatives fg.

It is important to realize that we are not calculating the derivative of the quotient fg.

Applying L’Hôpital’s Rule (0/0 Case)

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

  1. limx01cosxx
  2. limx1sin(πx)lnx
  3. limxe1/x11/x
  4. limx0sinxxx2
  1. Since the numerator 1cosx0

    and the denominator

    x0,

    we can apply L’Hôpital’s rule to evaluate this limit. We have

    limx01cosxx=limx0ddx(1cosx)ddx(x)=limx0sinx1=limx0(sinx)limx0(1)=01=0.

  2. As x1,

    the numerator

    sin(πx)0

    and the denominator

    ln(x)0.

    Therefore, we can apply L’Hôpital’s rule. We obtain

    limx1sin(πx)lnx=limx1πcos(πx)1/x=limx1(πx)cos(πx)=(π·1)(−1)=π.

  3. As x,

    the numerator

    e1/x10

    and the denominator

    (1x)0.

    Therefore, we can apply L’Hôpital’s rule. We obtain

    limxe1/x11x=limxe1/x(−1x2)(−1x2)=limxe1/x=e0=1.

  4. As x0,

    both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain

    limx0sinxxx2=limx0cosx12x.

    Since the numerator and denominator of this new quotient both approach zero as

    x0,

    we apply L’Hôpital’s rule again. In doing so, we see that

    limx0cosx12x=limx0sinx2=0.

    Therefore, we conclude that

    limx0sinxxx2=0.

Evaluate limx0xtanx.

1

Hint

ddxtanx=sec2x

We can also use L’Hôpital’s rule to evaluate limits of quotients f(x)g(x)

in which f(x)±

and 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 .

L’Hôpital’s Rule(/Case)

Suppose f

and g

are differentiable functions over an open interval containing a,

except possibly at a.

Suppose limxaf(x)=

(or )

and limxag(x)=

(or ).

Then,

limxaf(x)g(x)=limxaf(x)g(x),

assuming the limit on the right exists or is

or .

This result also holds if the limit is infinite, if a=

or ,

or the limit is one-sided.

Applying L’Hôpital’s Rule(/Case)

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

  1. limx3x+52x+1
  2. limx0+lnxcotx
  1. Since 3x+5

    and

    2x+1

    are first-degree polynomials with positive leading coefficients,

    limx(3x+5)=

    and

    limx(2x+1)=.

    Therefore, we apply L’Hôpital’s rule and obtain

    limx3x+52x+1=limx32=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

    x

    in the denominator. In doing so, we saw that

    limx3x+52x+1=limx3+5/x2x+1/x=32.

    L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.

  2. Here, limx0+lnx=

    and

    limx0+cotx=.

    Therefore, we can apply L’Hôpital’s rule and obtain

    limx0+lnxcotx=limx0+1/xcsc2x=limx0+1xcsc2x.

    Now as

    x0+, 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

    cscx

    to write

    limx0+1xcsc2x=limx0+sin2xx.

    Now

    limx0+sin2x=0

    and

    limx0+x=0,

    so we apply L’Hôpital’s rule again. We find

    limx0+sin2xx=limx0+2sinxcosx−1=0−1=0.

    We conclude that

    limx0+lnxcotx=0.

Evaluate limxlnx5x.

Hint

ddxlnx=1x

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),

it is essential that the limit of f(x)g(x)

be of the form 00

or /.

Consider the following example.

When L’Hôpital’s Rule Does Not Apply

Consider limx1x2+53x+4.

Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

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

ddx(x2+5)=2x

and

ddx(3x+4)=3.

At which point we would conclude erroneously that

limx1x2+53x+4=limx12x3=23.

However, since limx1(x2+5)=6

and limx1(3x+4)=7,

we actually have

limx1x2+53x+4=67.

We can conclude that

limx1x2+53x+4limx1ddx(x2+5)ddx(3x+4).

Explain why we cannot apply L’Hôpital’s rule to evaluate limx0+cosxx.

Evaluate limx0+cosxx

by other means.

limx0+cosx=1.

Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is

Hint

Determine the limits of the numerator and denominator separately.

Other Indeterminate Forms

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms 00

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·,

,1,0,

and 00

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 00

or /.

Indeterminate Form of Type 0·

Suppose we want to evaluate limxa(f(x)·g(x)),

where f(x)0

and g(x)

(or )

as xa.

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·

to denote the form that arises in this situation. The expression 0·

is considered indeterminate because we cannot determine without further analysis the exact behavior of the product f(x)g(x)

as xa.

For example, let n

be a positive integer and consider

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

As x,

f(x)0

and g(x).

However, the limit as x

of f(x)g(x)=3x2(xn+1)

varies, depending on n.

If n=2,

then limxf(x)g(x)=3.

If n=1,

then limxf(x)g(x)=.

If n=3,

then limxf(x)g(x)=0.

Here we consider another limit involving the indeterminate form 0·

and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

Indeterminate Form of Type0·

Evaluate limx0+xlnx.

First, rewrite the function xlnx

as a quotient to apply L’Hôpital’s rule. If we write

xlnx=lnx1/x,

we see that lnx

as x0+

and 1x

as x0+.

Therefore, we can apply L’Hôpital’s rule and obtain

limx0+lnx1/x=limx0+ddx(lnx)ddx(1/x)=limx0+1/x-1/x2=limx0+(x)=0.

We conclude that

limx0+xlnx=0.

L’Hôpital’s Rule · Calculus (2)

Evaluate limx0xcotx.

1

Hint

Write xcotx=xcosxsinx

Indeterminate Form of Type

Another type of indeterminate form is .

Consider the following example. Let n

be a positive integer and let f(x)=3xn

and g(x)=3x2+5.

As x,

f(x)

and g(x).

We are interested in limx(f(x)g(x)).

Depending on whether f(x)

grows faster, g(x)

grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since f(x)

and 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 n

in the function f(x)=3xn

is n=3,

then

limx(f(x)g(x))=limx(3x33x25)=.

On the other hand, if n=2,

then

limx(f(x)g(x))=limx(3x23x25)=−5.

However, if n=1,

then

limx(f(x)g(x))=limx(3x3x25)=.

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.

Indeterminate Form of Type

Evaluate limx0+(1x21tanx).

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

we have

1x21tanx=(tanx)x2x2tanx.

As x0+,

the numerator tanxx20

and the denominator x2tanx0.

Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

limx0+(tanx)x2x2tanx=limx0+(sec2x)2xx2sec2x+2xtanx.

As x0+,

(sec2x)2x1

and x2sec2x+2xtanx0.

Since the denominator is positive as x

approaches zero from the right, we conclude that

limx0+(sec2x)2xx2sec2x+2xtanx=.

Therefore,

limx0+(1x21tanx)=.

Evaluate limx0+(1x1sinx).

Hint

Rewrite the difference of fractions as a single fraction.

Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions 00,

0,

and 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 limxaf(x)g(x)

and we arrive at the indeterminate form 0.

(The indeterminate forms 00

and 1

can be handled similarly.) We proceed as follows. Let

y=f(x)g(x).

Then,

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

Therefore,

limxa[ln(y)]=limxa[g(x)ln(f(x))].

Since limxaf(x)=,

we know that limxaln(f(x))=.

Therefore, limxag(x)ln(f(x))

is of the indeterminate form 0·,

and we can use the techniques discussed earlier to rewrite the expression g(x)ln(f(x))

in a form so that we can apply L’Hôpital’s rule. Suppose limxag(x)ln(f(x))=L,

where L

may be

or .

Then

limxa[ln(y)]=L.

Since the natural logarithm function is continuous, we conclude that

ln(limxay)=L,

which gives us

limxay=limxaf(x)g(x)=eL.

Indeterminate Form of Type0

Evaluate limxx1/x.

Let y=x1/x.

Then,

ln(x1/x)=1xlnx=lnxx.

We need to evaluate limxlnxx.

Applying L’Hôpital’s rule, we obtain

limxlny=limxlnxx=limx1/x1=0.

Therefore, limxlny=0.

Since the natural logarithm function is continuous, we conclude that

ln(limxy)=0,

which leads to

limxy=limxlnxx=e0=1.

Hence,

limxx1/x=1.

Evaluate limxx1/ln(x).

e

Hint

Let y=x1/ln(x)

and apply the natural logarithm to both sides of the equation.

Indeterminate Form of Type00

Evaluate limx0+xsinx.

Let

y=xsinx.

Therefore,

lny=ln(xsinx)=sinxlnx.

We now evaluate limx0+sinxlnx.

Since limx0+sinx=0

and limx0+lnx=,

we have the indeterminate form 0·.

To apply L’Hôpital’s rule, we need to rewrite sinxlnx

as a fraction. We could write

sinxlnx=sinx1/lnx

or

sinxlnx=lnx1/sinx=lnxcscx.

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

limx0+sinxlnx=limx0+sinx1/lnx=limx0+cosx−1/(x(lnx)2)=limx0+(x(lnx)2cosx).

Unfortunately, we not only have another expression involving the indeterminate form 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

sinxlnx=lnx1/sinx=lnxcscx,

and applying L’Hôpital’s rule, we obtain

limx0+sinxlnx=limx0+lnxcscx=limx0+1/xcscxcotx=limx0+−1xcscxcotx.

Using the fact that cscx=1sinx

and cotx=cosxsinx,

we can rewrite the expression on the right-hand side as

limx0+sin2xxcosx=limx0+[sinxx·(tanx)]=(limx0+sinxx)·(limx0+(tanx))=1·0=0.

We conclude that limx0+lny=0.

Therefore, ln(limx0+y)=0

and we have

limx0+y=limx0+xsinx=e0=1.

Hence,

limx0+xsinx=1.

Evaluate limx0+xx.

1

Hint

Let y=xx

and take the natural logarithm of both sides of the equation.

Growth Rates of Functions

Suppose the functions f

and g

both approach infinity as x.

Although the values of both functions become arbitrarily large as the values of x

become sufficiently large, sometimes one function is growing more quickly than the other. For example, f(x)=x2

and g(x)=x3

both approach infinity as x.

However, as shown in the following table, the values of x3

are growing much faster than the values of x2.

Comparing the Growth Rates of x2 and x3
x10100100010,000
f(x)=x210010,0001,000,000100,000,000
g(x)=x310001,000,0001,000,000,0001,000,000,000,000

In fact,

limxx3x2=limxx=.or, equivalently,limxx2x3=limx1x=0.

As a result, we say x3

is growing more rapidly than x2

as x.

On the other hand, for f(x)=x2

and g(x)=3x2+4x+1,

although the values of g(x)

are always greater than the values of f(x)

for x>0,

each value of g(x)

is roughly three times the corresponding value of f(x)

as x,

as shown in the following table. In fact,

limxx23x2+4x+1=13.

Comparing the Growth Rates of x2 and 3x2+4x+1
x10100100010,000
f(x)=x210010,0001,000,000100,000,000
g(x)=3x2+4x+134130,4013,004,001300,040,001

In this case, we say that x2

and 3x2+4x+1

are growing at the same rate as x.

More generally, suppose f

and g

are two functions that approach infinity as x.

We say g

grows more rapidly than f

as x

if

limxg(x)f(x)=;or, equivalently,limxf(x)g(x)=0.

On the other hand, if there exists a constant M0

such that

limxf(x)g(x)=M,

we say f

and g

grow at the same rate as x.

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

Comparing the Growth Rates ofln(x),x2,andex

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

  1. f(x)=x2andg(x)=ex
  2. f(x)=ln(x)andg(x)=x2
  1. Since limxx2=

    and

    limxex=,

    we can use L’Hôpital’s rule to evaluate

    limx[x2ex].

    We obtain

    limxx2ex=limx2xex.

    Since

    limx2x=

    and

    limxex=,

    we can apply L’Hôpital’s rule again. Since

    limx2xex=limx2ex=0,

    we conclude that

    limxx2ex=0.

    Therefore,

    ex

    grows more rapidly than

    x2

    as

    x

    (See [link] and [link]).

    L’Hôpital’s Rule · Calculus (3)

    Growth rates of a power function and an exponential function.
    x5101520
    x225100225400
    ex14822,0263,269,017485,165,195
  2. Since limxlnx=

    and

    limxx2=,

    we can use L’Hôpital’s rule to evaluate

    limxlnxx2.

    We obtain

    limxlnxx2=limx1/x2x=limx12x2=0.

    Thus,

    x2

    grows more rapidly than

    lnx

    as

    x

    (see [link] and [link]).

    L’Hôpital’s Rule · Calculus (4)

    Growth rates of a power function and a logarithmic function
    x10100100010,000
    ln(x)2.3034.6056.9089.210
    x210010,0001,000,000100,000,000

Compare the growth rates of x100

and 2x.

The function 2x

grows faster than x100.

Hint

Apply L’Hôpital’s rule to x100/2x

Using the same ideas as in [link]a. it is not difficult to show that ex

grows more rapidly than xp

for any p>0.

In [link] and [link], we compare ex

with x3

and x4

as x.

L’Hôpital’s Rule · Calculus (5)

An exponential function grows at a faster rate than any power function
x5101520
x3125100033758000
x462510,00050,625160,000
ex14822,0263,269,017485,165,195

Similarly, it is not difficult to show that xp

grows more rapidly than lnx

for any p>0.

In [link] and [link], we compare lnx

with x3

and x.

L’Hôpital’s Rule · Calculus (6)

A logarithmic function grows at a slower rate than any root function
x10100100010,000
ln(x)2.3034.6056.9089.210
x32.1544.6421021.544
x3.1621031.623100

Key Concepts

  • L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form 00

    or

    /

    arises.

  • L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form 00

    or

    /.
  • The exponential function ex

    grows faster than any power function

    xp, p>0.
  • The logarithmic function lnx

    grows more slowly than any power function

    xp, p>0.

For the following exercises, evaluate the limit.

Evaluate the limit limxexx.

Evaluate the limit limxexxk.

Evaluate the limit limxlnxxk.

Evaluate the limit limxaxax2a2,a0

.

12a

Evaluate the limit limxaxax3a3,a0

.

Evaluate the limit limxaxaxnan,a0

.

1nan1

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.

limx0+x2lnx

limxx1/x

Cannot apply directly; use logarithms

limx0x2/x

limx0x21/x

Cannot apply directly; rewrite as limx0x3

limxexx

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

limx3x29x3

6

limx3x29x+3

limx0(1+x)−21x

−2

limxπ/2cosxπ2x

limxπxπsinx

−1

limx1x1sinx

limx0(1+x)n1x

n

limx0(1+x)n1nxx2

limx0sinxtanxx3

12

limx01+x1xx

limx0exx1x2

12

limx0tanxx

limx1x1lnx

1

limx0(x+1)1/x

limx1xx3x1

16

limx0+x2x

limxxsin(1x)

1

limx0sinxxx2

limx0+xln(x4)

limx(xex)

limxx2ex

limx03x2xx

limx01+1/x11/x

−1

limxπ/4(1tanx)cotx

limxxe1/x

limx0x1/cosx

limx0x1/x

1

limx0(11x)x

limx(11x)x

1e

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.

[T] limx0ex1x

[T] limx0xsin(1x)

[T] limx1x11cos(πx)

[T] limx1e(x1)1x1

1

[T] limx1(x1)2lnx

[T] limxπ1+cosxsinx

[T] limx0(cscx1x)

[T] limx0+tan(xx)

tan(1)

[T] limx0+lnxsinx

[T] limx0exexx

2

Glossary

indeterminate forms
when evaluating a limit, the forms 00, /, 0·, , 00, 0,

and

1

are considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its value is

L’Hôpital’s rule
if f

and

g

are differentiable functions over an interval

a,

except possibly at

a,

and

limxaf(x)=0=limxag(x)

or

limxaf(x)

and

limxag(x)

are infinite, then

limxaf(x)g(x)=limxaf(x)g(x),

assuming the limit on the right exists or is

or

L’Hôpital’s Rule · Calculus (7)
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L’Hôpital’s Rule · Calculus (2024)

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