L’Hôpital’s Rule

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

This is considered an indeterminate form because we cannot determine the exact behavior of

\frac{f (x)}{g (x)}

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

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

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

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

\frac{0}{0} .

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

\frac{0}{0}

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 $lim_{x \to a} f (x) = 0$

and $lim_{x \to a} g (x) = 0,$

then

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)},

assuming the limit on the right exists or is $\infty$

or $− \infty .$

This result also holds if we are considering one-sided limits, or if $a = \infty and - \infty .$

Proof

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

\frac{f}{g}

by considering the limit of the quotient of the derivatives

\frac{f^{'}}{g^{'}} .

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

\frac{f}{g} .

Applying L’Hôpital’s Rule (0/0 Case)

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

$lim_{x \to 0} \frac{1 - cos x}{x}$
$lim_{x \to 1} \frac{sin (π x)}{ln x}$
$lim_{x \to \infty} \frac{e^{1 / x} - 1}{1 / x}$
$lim_{x \to 0} \frac{sin x - x}{x^{2}}$

Since the numerator $1 - cos x \to 0$
and the denominator
$x \to 0,$
we can apply L’Hôpital’s rule to evaluate this limit. We have

$\begin{matrix} lim_{x \to 0} \frac{1 - cos x}{x} & = lim_{x \to 0} \frac{\frac{d}{d x} (1 - cos x)}{\frac{d}{d x} (x)} \\ = lim_{x \to 0} \frac{sin x}{1} \\ = \frac{lim_{x \to 0} (sin x)}{lim_{x \to 0} (1)} \\ = \frac{0}{1} = 0. \end{matrix}$
As $x \to 1,$
the numerator
$sin (π x) \to 0$
and the denominator
$ln (x) \to 0 .$
Therefore, we can apply L’Hôpital’s rule. We obtain

$\begin{matrix} lim_{x \to 1} \frac{sin (π x)}{ln x} & = lim_{x \to 1} \frac{π cos (π x)}{1 / x} \\ = lim_{x \to 1} (π x) cos (π x) \\ = (π \cdot 1) (−1) = − π . \end{matrix}$
As $x \to \infty,$
the numerator
$e^{1 / x} - 1 \to 0$
and the denominator
$(\frac{1}{x}) \to 0 .$
Therefore, we can apply L’Hôpital’s rule. We obtain

$lim_{x \to \infty} \frac{e^{1 / x} - 1}{\frac{1}{x}} = lim_{x \to \infty} \frac{e^{1 / x} (\frac{−1}{x^{2}})}{(\frac{−1}{x^{2}})} = lim_{x \to \infty} e^{1 / x} = e^{0} = 1 .$
As $x \to 0,$
both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain

$lim_{x \to 0} \frac{sin x - x}{x^{2}} = lim_{x \to 0} \frac{cos x - 1}{2 x} .$

Since the numerator and denominator of this new quotient both approach zero as
$x \to 0,$
we apply L’Hôpital’s rule again. In doing so, we see that

$lim_{x \to 0} \frac{cos x - 1}{2 x} = lim_{x \to 0} \frac{− sin x}{2} = 0 .$

Therefore, we conclude that

$lim_{x \to 0} \frac{sin x - x}{x^{2}} = 0 .$

We can also use L’Hôpital’s rule to evaluate limits of quotients

\frac{f (x)}{g (x)}

Limits of this form are classified as indeterminate forms of type

\infty / \infty .

L’Hôpital’s Rule

(\infty / \infty

Case)

Suppose $f$

and $g$

are differentiable functions over an open interval containing $a,$

except possibly at $a .$

Suppose $lim_{x \to a} f (x) = \infty$

(or $− \infty)$

and $lim_{x \to a} g (x) = \infty$

(or $− \infty) .$

Then,

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)},

assuming the limit on the right exists or is $\infty$

or $− \infty .$

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

or $− \infty,$

or the limit is one-sided.

Applying L’Hôpital’s Rule

(\infty / \infty

Case)

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

$lim_{x \to \infty} \frac{3 x + 5}{2 x + 1}$
$lim_{x \to 0^{+}} \frac{ln x}{cot x}$

Since $3 x + 5$
and
$2 x + 1$
are first-degree polynomials with positive leading coefficients,
$lim_{x \to \infty} (3 x + 5) = \infty$
and
$lim_{x \to \infty} (2 x + 1) = \infty .$
Therefore, we apply L’Hôpital’s rule and obtain

$lim_{x \to \infty} \frac{3 x + 5}{2 x + 1} = lim_{x \to \infty} \frac{3}{2} = \frac{3}{2} .$

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

$lim_{x \to \infty} \frac{3 x + 5}{2 x + 1} = lim_{x \to \infty} \frac{3 + 5 / x}{2 x + 1 / x} = \frac{3}{2} .$

L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
Here, $lim_{x \to 0^{+}} ln x = − \infty$
and
$lim_{x \to 0^{+}} cot x = \infty .$
Therefore, we can apply L’Hôpital’s rule and obtain

$lim_{x \to 0^{+}} \frac{ln x}{cot x} = lim_{x \to 0^{+}} \frac{1 / x}{− {csc}^{2} x} = lim_{x \to 0^{+}} \frac{1}{− x {csc}^{2} x} .$

Now as
$x \to 0^{+},$ ${csc}^{2} x \to \infty .$
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
$csc x$
to write

$lim_{x \to 0^{+}} \frac{1}{− x {csc}^{2} x} = lim_{x \to 0^{+}} \frac{{sin}^{2} x}{− x} .$

Now
$lim_{x \to 0^{+}} {sin}^{2} x = 0$
and
$lim_{x \to 0^{+}} x = 0,$
so we apply L’Hôpital’s rule again. We find

$lim_{x \to 0^{+}} \frac{{sin}^{2} x}{− x} = lim_{x \to 0^{+}} \frac{2 sin x cos x}{−1} = \frac{0}{−1} = 0 .$

We conclude that

$lim_{x \to 0^{+}} \frac{ln x}{cot x} = 0 .$

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

\frac{f (x)}{g (x)},

When L’Hôpital’s Rule Does Not Apply

Consider $lim_{x \to 1} \frac{x^{2} + 5}{3 x + 4} .$

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

\frac{d}{d x} (x^{2} + 5) = 2 x

and

\frac{d}{d x} (3 x + 4) = 3 .

At which point we would conclude erroneously that

lim_{x \to 1} \frac{x^{2} + 5}{3 x + 4} = lim_{x \to 1} \frac{2 x}{3} = \frac{2}{3} .

However, since $lim_{x \to 1} (x^{2} + 5) = 6$

and $lim_{x \to 1} (3 x + 4) = 7,$

we actually have

lim_{x \to 1} \frac{x^{2} + 5}{3 x + 4} = \frac{6}{7} .

We can conclude that

lim_{x \to 1} \frac{x^{2} + 5}{3 x + 4} \neq lim_{x \to 1} \frac{\frac{d}{d x} (x^{2} + 5)}{\frac{d}{d x} (3 x + 4)} .

Other Indeterminate Forms

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

\frac{0}{0}

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 \cdot \infty,

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

\frac{0}{0}

Indeterminate Form of Type

0 \cdot \infty

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 \cdot \infty

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

0 \cdot \infty

is considered indeterminate because we cannot determine without further analysis the exact behavior of the product

f (x) g (x)

Here we consider another limit involving the indeterminate form

0 \cdot \infty

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

Indeterminate Form of Type

0 \cdot \infty

Evaluate $lim_{x \to 0^{+}} x ln x .$

First, rewrite the function $x ln x$

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

x ln x = \frac{ln x}{1 / x},

we see that $ln x \to − \infty$

as $x \to 0^{+}$

and $\frac{1}{x} \to \infty$

as $x \to 0^{+} .$

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

lim_{x \to 0^{+}} \frac{ln x}{1 / x} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} (ln x)}{\frac{d}{d x} (1 / x)} = lim_{x \to 0^{+}} \frac{1 / x}{-1 / x^{2}} = lim_{x \to 0^{+}} (− x) = 0 .

We conclude that

lim_{x \to 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.

Indeterminate Form of Type

\infty - \infty

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

f (x) \to \infty

to denote the form of this limit. As with our other indeterminate forms,

\infty - \infty

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

Therefore, the limit cannot be determined by considering only

\infty - \infty .

Next we see how to rewrite an expression involving the indeterminate form

\infty - \infty

Indeterminate Form of Type

\infty - \infty

Evaluate $lim_{x \to 0^{+}} (\frac{1}{x^{2}} - \frac{1}{tan x}) .$

By combining the fractions, we can write the function as a quotient. Since the least common denominator is $x^{2} tan x,$

we have

\frac{1}{x^{2}} - \frac{1}{tan x} = \frac{(tan x) - x^{2}}{x^{2} tan x} .

As $x \to 0^{+},$

the numerator $tan x - x^{2} \to 0$

and the denominator $x^{2} tan x \to 0 .$

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

lim_{x \to 0^{+}} \frac{(tan x) - x^{2}}{x^{2} tan x} = lim_{x \to 0^{+}} \frac{({sec}^{2} x) - 2 x}{x^{2} {sec}^{2} x + 2 x tan x} .

As $x \to 0^{+},$

({sec}^{2} x) - 2 x \to 1

and $x^{2} {sec}^{2} x + 2 x tan x \to 0 .$

Since the denominator is positive as $x$

approaches zero from the right, we conclude that

lim_{x \to 0^{+}} \frac{({sec}^{2} x) - 2 x}{x^{2} {sec}^{2} x + 2 x tan x} = \infty .

Therefore,

lim_{x \to 0^{+}} (\frac{1}{x^{2}} - \frac{1}{tan x}) = \infty .

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

0^{0},

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

lim_{x \to a} f {(x)}^{g (x)}

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

lim_{x \to a} g (x) ln (f (x)) = L,

Indeterminate Form of Type

\infty^{0}

Evaluate $lim_{x \to \infty} x^{1 / x} .$

Let $y = x^{1 / x} .$

Then,

ln (x^{1 / x}) = \frac{1}{x} ln x = \frac{ln x}{x} .

We need to evaluate $lim_{x \to \infty} \frac{ln x}{x} .$

Applying L’Hôpital’s rule, we obtain

lim_{x \to \infty} ln y = lim_{x \to \infty} \frac{ln x}{x} = lim_{x \to \infty} \frac{1 / x}{1} = 0 .

Therefore, $lim_{x \to \infty} ln y = 0 .$

Since the natural logarithm function is continuous, we conclude that

ln (lim_{x \to \infty} y) = 0,

which leads to

lim_{x \to \infty} y = lim_{x \to \infty} \frac{ln x}{x} = e^{0} = 1 .

Hence,

lim_{x \to \infty} x^{1 / x} = 1 .

Indeterminate Form of Type

0^{0}

Evaluate $lim_{x \to 0^{+}} x^{sin x} .$

Let

y = x^{sin x} .

Therefore,

ln y = ln (x^{sin x}) = sin x ln x .

We now evaluate $lim_{x \to 0^{+}} sin x ln x .$

Since $lim_{x \to 0^{+}} sin x = 0$

and $lim_{x \to 0^{+}} ln x = − \infty,$

we have the indeterminate form $0 \cdot \infty .$

To apply L’Hôpital’s rule, we need to rewrite $sin x ln x$

as a fraction. We could write

sin x ln x = \frac{sin x}{1 / ln x}

sin x ln x = \frac{ln x}{1 / sin x} = \frac{ln x}{csc x} .

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

lim_{x \to 0^{+}} sin x ln x = lim_{x \to 0^{+}} \frac{sin x}{1 / ln x} = lim_{x \to 0^{+}} \frac{cos x}{−1 / (x {(ln x)}^{2})} = lim_{x \to 0^{+}} (− x {(ln x)}^{2} cos x) .

Unfortunately, we not only have another expression involving the indeterminate form $0 \cdot \infty,$

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 = \frac{ln x}{1 / sin x} = \frac{ln x}{csc x},

and applying L’Hôpital’s rule, we obtain

lim_{x \to 0^{+}} sin x ln x = lim_{x \to 0^{+}} \frac{ln x}{csc x} = lim_{x \to 0^{+}} \frac{1 / x}{− csc x cot x} = lim_{x \to 0^{+}} \frac{−1}{x csc x cot x} .

Using the fact that $csc x = \frac{1}{sin x}$

and $cot x = \frac{cos x}{sin x},$

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

lim_{x \to 0^{+}} \frac{− {sin}^{2} x}{x cos x} = lim_{x \to 0^{+}} [\frac{sin x}{x} \cdot (− tan x)] = (lim_{x \to 0^{+}} \frac{sin x}{x}) \cdot (lim_{x \to 0^{+}} (− tan x)) = 1 \cdot 0 = 0 .

We conclude that $lim_{x \to 0^{+}} ln y = 0 .$

Therefore, $ln (lim_{x \to 0^{+}} y) = 0$

and we have

lim_{x \to 0^{+}} y = lim_{x \to 0^{+}} x^{sin x} = e^{0} = 1 .

Hence,

lim_{x \to 0^{+}} x^{sin x} = 1 .

Growth Rates of Functions

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) = x^{2}

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 of $x^{2}$ and $x^{3}$
$x$	$10$	$100$	$1000$	$10,000$
$f (x) = x^{2}$	$100$	$10,000$	$1,000,000$	$100,000,000$
$g (x) = x^{3}$	$1000$	$1,000,000$	$1,000,000,000$	$1,000,000,000,000$

Comparing the Growth Rates of $x^{2}$ and $3 x^{2} + 4 x + 1$
$x$	$10$	$100$	$1000$	$10,000$
$f (x) = x^{2}$	$100$	$10,000$	$1,000,000$	$100,000,000$
$g (x) = 3 x^{2} + 4 x + 1$	$341$	$30,401$	$3,004,001$	$300,040,001$

Comparing the Growth Rates of

ln (x),

x^{2},

and

e^{x}

For each of the following pairs of functions, use L’Hôpital’s rule to evaluate $lim_{x \to \infty} (\frac{f (x)}{g (x)}) .$

$f (x) = x^{2} and g (x) = e^{x}$
$f (x) = ln (x) and g (x) = x^{2}$

Since $lim_{x \to \infty} x^{2} = \infty$
and
$lim_{x \to \infty} e^{x} = \infty,$
we can use L’Hôpital’s rule to evaluate
$lim_{x \to \infty} [\frac{x^{2}}{e^{x}}] .$
We obtain

$lim_{x \to \infty} \frac{x^{2}}{e^{x}} = lim_{x \to \infty} \frac{2 x}{e^{x}} .$

Since
$lim_{x \to \infty} 2 x = \infty$
and
$lim_{x \to \infty} e^{x} = \infty,$
we can apply L’Hôpital’s rule again. Since

$lim_{x \to \infty} \frac{2 x}{e^{x}} = lim_{x \to \infty} \frac{2}{e^{x}} = 0,$

we conclude that

$lim_{x \to \infty} \frac{x^{2}}{e^{x}} = 0 .$

Therefore,
$e^{x}$
grows more rapidly than
$x^{2}$
as
$x \to \infty$
(See [link] and [link]).

Growth rates of a power function and an exponential function.

$x$ $5$ $10$ $15$ $20$

$x^{2}$ $25$ $100$ $225$ $400$

$e^{x}$ $148$ $22,026$ $3,269,017$ $485,165,195$
Since $lim_{x \to \infty} ln x = \infty$
and
$lim_{x \to \infty} x^{2} = \infty,$
we can use L’Hôpital’s rule to evaluate
$lim_{x \to \infty} \frac{ln x}{x^{2}} .$
We obtain

$lim_{x \to \infty} \frac{ln x}{x^{2}} = lim_{x \to \infty} \frac{1 / x}{2 x} = lim_{x \to \infty} \frac{1}{2 x^{2}} = 0 .$

Thus,
$x^{2}$
grows more rapidly than
$ln x$
as
$x \to \infty$
(see [link] and [link]).

Growth rates of a power function and a logarithmic function

$x$ $10$ $100$ $1000$ $10,000$

$ln (x)$ $2.303$ $4.605$ $6.908$ $9.210$

$x^{2}$ $100$ $10,000$ $1,000,000$ $100,000,000$

Key Concepts

Growth rates of a power function and an exponential function.
$x$	$5$	$10$	$15$	$20$
$x^{2}$	$25$	$100$	$225$	$400$
$e^{x}$	$148$	$22,026$	$3,269,017$	$485,165,195$

Growth rates of a power function and a logarithmic function
$x$	$10$	$100$	$1000$	$10,000$
$ln (x)$	$2.303$	$4.605$	$6.908$	$9.210$
$x^{2}$	$100$	$10,000$	$1,000,000$	$100,000,000$

An exponential function grows at a faster rate than any power function
$x$	$5$	$10$	$15$	$20$
$x^{3}$	$125$	$1000$	$3375$	$8000$
$x^{4}$	$625$	$10,000$	$50,625$	$160,000$
$e^{x}$	$148$	$22,026$	$3,269,017$	$485,165,195$

A logarithmic function grows at a slower rate than any root function
$x$	$10$	$100$	$1000$	$10,000$
$ln (x)$	$2.303$	$4.605$	$6.908$	$9.210$
$\sqrt[3]{x}$	$2.154$	$4.642$	$10$	$21.544$
$\sqrt{x}$	$3.162$	$10$	$31.623$	$100$

For the following exercises, evaluate the limit.

Evaluate the limit $lim_{x \to \infty} \frac{e^{x}}{x} .$

Evaluate the limit $lim_{x \to \infty} \frac{e^{x}}{x^{k}} .$

\infty

Evaluate the limit $lim_{x \to \infty} \frac{ln x}{x^{k}} .$

Evaluate the limit $lim_{x \to a} \frac{x - a}{x^{2} - a^{2}}, a \neq 0$

\frac{1}{2 a}

Evaluate the limit $lim_{x \to a} \frac{x - a}{x^{3} - a^{3}}, a \neq 0$

Evaluate the limit $lim_{x \to a} \frac{x - a}{x^{n} - a^{n}}, a \neq 0$

\frac{1}{n a^{n - 1}}

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.

lim_{x \to 0^{+}} x^{2} ln x

lim_{x \to \infty} x^{1 / x}

Cannot apply directly; use logarithms

lim_{x \to 0} x^{2 / x}

lim_{x \to 0} \frac{x^{2}}{1 / x}

Cannot apply directly; rewrite as $lim_{x \to 0} x^{3}$

lim_{x \to \infty} \frac{e^{x}}{x}

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

lim_{x \to 3} \frac{x^{2} - 9}{x - 3}

6

lim_{x \to 3} \frac{x^{2} - 9}{x + 3}

lim_{x \to 0} \frac{{(1 + x)}^{−2} - 1}{x}

−2

lim_{x \to π / 2} \frac{cos x}{\frac{π}{2} - x}

lim_{x \to π} \frac{x - π}{sin x}

−1

lim_{x \to 1} \frac{x - 1}{sin x}

lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{x}

n

lim_{x \to 0} \frac{{(1 + x)}^{n} - 1 - n x}{x^{2}}

lim_{x \to 0} \frac{sin x - tan x}{x^{3}}

- \frac{1}{2}

lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}

lim_{x \to 0} \frac{e^{x} - x - 1}{x^{2}}

\frac{1}{2}

lim_{x \to 0} \frac{tan x}{\sqrt{x}}

lim_{x \to 1} \frac{x - 1}{ln x}

1

lim_{x \to 0} {(x + 1)}^{1 / x}

lim_{x \to 1} \frac{\sqrt{x} - \sqrt[3]{x}}{x - 1}

\frac{1}{6}

lim_{x \to 0^{+}} x^{2 x}

lim_{x \to \infty} x sin (\frac{1}{x})

1

lim_{x \to 0} \frac{sin x - x}{x^{2}}

lim_{x \to 0^{+}} x ln (x^{4})

0

lim_{x \to \infty} (x - e^{x})

lim_{x \to \infty} x^{2} e^{− x}

0

lim_{x \to 0} \frac{3^{x} - 2^{x}}{x}

lim_{x \to 0} \frac{1 + 1 / x}{1 - 1 / x}

−1

lim_{x \to π / 4} (1 - tan x) cot x

lim_{x \to \infty} x e^{1 / x}

\infty

lim_{x \to 0} x^{1 / cos x}

lim_{x \to 0} x^{1 / x}

1

lim_{x \to 0} {(1 - \frac{1}{x})}^{x}

lim_{x \to \infty} {(1 - \frac{1}{x})}^{x}

\frac{1}{e}

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] $lim_{x \to 0} \frac{e^{x} - 1}{x}$

[T] $lim_{x \to 0} x sin (\frac{1}{x})$

0

[T] $lim_{x \to 1} \frac{x - 1}{1 - cos (π x)}$

[T] $lim_{x \to 1} \frac{e^{(x - 1)} - 1}{x - 1}$

1

[T] $lim_{x \to 1} \frac{{(x - 1)}^{2}}{ln x}$

[T] $lim_{x \to π} \frac{1 + cos x}{sin x}$

0

[T] $lim_{x \to 0} (csc x - \frac{1}{x})$

[T] $lim_{x \to 0^{+}} tan (x^{x})$

tan (1)

[T] $lim_{x \to 0^{+}} \frac{ln x}{sin x}$

[T] $lim_{x \to 0} \frac{e^{x} - e^{− x}}{x}$

2

Glossary

indeterminate forms

when evaluating a limit, the forms

\frac{0}{0},

\infty / \infty,

0 \cdot \infty,

\infty - \infty,

0^{0},

\infty^{0},

and

1^{\infty}

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

f

and

g

are differentiable functions over an interval

a,

except possibly at

a,

and

lim_{x \to a} f (x) = 0 = lim_{x \to a} g (x)

lim_{x \to a} f (x)

and

lim_{x \to a} g (x)

are infinite, then

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)},

assuming the limit on the right exists or is

\infty

− \infty

L’Hôpital’s Rule

Applying L’Hôpital’s Rule

Proof

Other Indeterminate Forms

Indeterminate Form of Type 0·∞

Indeterminate Form of Type ∞−∞

Growth Rates of Functions

Key Concepts

Glossary

Indeterminate Form of Type $0 \cdot \infty$

Indeterminate Form of Type $\infty - \infty$