While this chapter is devoted to learning techniques of integration, this section is not about integration. Rather, it is concerned with a technique of evaluating certain limits that will be useful in the following section, where integration is once more discussed.

Our treatment of limits exposed us to “0/0”, an indeterminate form. If $\lim\limits_{x\to c}f(x)=0$ and $\lim\limits_{x\to c} g(x) =0\text{,}$ we do not conclude that $\lim\limits_{x\to c} f(x)/g(x)$ is $0/0\text{;}$ rather, we use $0/0$ as notation to describe the fact that both the numerator and denominator approach 0. The expression 0/0 has no numeric value; other work must be done to evaluate the limit.

Other indeterminate forms exist; they are: $\infty/\infty\text{,}$ $0\cdot\infty\text{,}$ $\infty-\infty\text{,}$ $0^0\text{,}$ $1^\infty$ and $\infty^0\text{.}$ Just as “0/0” does not mean “divide 0 by 0,” the expression “$\infty/\infty$” does not mean “divide infinity by infinity.” Instead, it means “a quantity is growing without bound and is being divided by another quantity that is growing without bound.” We cannot determine from such a statement what value, if any, results in the limit. Likewise, “$0\cdot \infty$” does not mean “multiply zero by infinity.” Instead, it means “one quantity is shrinking to zero, and is being multiplied by a quantity that is growing without bound.” We cannot determine from such a description what the result of such a limit will be.

This section introduces l'Hôpital's Rule, a method of resolving limits that produce the indeterminate forms 0/0 and $\infty/\infty\text{.}$ We'll also show how algebraic manipulation can be used to convert other indeterminate expressions into one of these two forms so that our new rule can be applied.

We demonstrate the use of l'Hôpital's Rule in the following examples; we will often use “LHR” as an abbreviation of “l'Hôpital's Rule.”

##### Example6.7.2Using l'Hôpital's Rule

Evaluate the following limits, using l'Hôpital's Rule as needed.

1. $\lim\limits_{x\to0}\frac{\sin(x) }x$

2. $\lim\limits_{x\to 1}\frac{\sqrt{x+3}-2}{1-x}$

3. $\lim\limits_{x\to0}\frac{x^2}{1-\cos(x) }$

4. $\lim\limits_{x\to 2}\frac{x^2+x-6}{x^2-3x+2}$

Solution

Note that at each step where l'Hôpital's Rule was applied, it was needed: the initial limit returned the indeterminate form of “$0/0\text{.}$” If the initial limit returns, for example, 1/2, then l'Hôpital's Rule does not apply.

The following theorem extends our initial version of l'Hôpital's Rule in two ways. It allows the technique to be applied to the indeterminate form $\infty/\infty$ and to limits where $x$ approaches $\pm\infty\text{.}$

##### Example6.7.4Using l'Hôpital's Rule with limits involving $\infty$

Evaluate the following limits.

$\ds 1.\ \lim_{x\to\infty} \frac{3x^2-100x+2}{4x^2+5x-1000} \qquad\qquad 2. \ \lim_{x\to \infty}\frac{e^x}{x^3}\text{.}$

Solution

# Subsection6.7.1Indeterminate Forms $0\cdot\infty$ and $\infty-\infty$

L'Hôpital's Rule can only be applied to ratios of functions. When faced with an indeterminate form such as $0\cdot\infty$ or $\infty-\infty\text{,}$ we can sometimes apply algebra to rewrite the limit so that l'Hôpital's Rule can be applied. We demonstrate the general idea in the next example.

##### Example6.7.5Applying l'Hôpital's Rule to other indeterminate forms

Evaluate the following limits.

1. $\lim\limits_{x\to0^+} x\cdot e^{1/x}$

2. $\lim\limits_{x\to0^-} x\cdot e^{1/x}$

3. $\lim\limits_{x\to\infty} \ln(x+1)-\ln(x)$

4. $\lim\limits_{x\to\infty} x^2-e^x$

Solution

# Subsection6.7.2Indeterminate Forms  $0^0\text{,}$ $1^\infty$ and $\infty^0$

When faced with an indeterminate form that involves a power, it often helps to employ the natural logarithmic function. The following Key Idea expresses the concept, which is followed by an example that demonstrates its use.

##### Key Idea6.7.6Evaluating Limits Involving Indeterminate Forms $0^0\text{,}$ $1^\infty$ and $\infty^0$

If $\lim\limits_{x\to c} \ln\big(f(x)\big) = L\text{,}$then $\lim\limits_{x\to c} f(x) = \lim_{x\to c} e^{\ln(f(x))} = e^L\text{.}$

##### Example6.7.7Using l'Hôpital's Rule with indeterminate forms involving exponents

Evaluate the following limits.

$\ds 1. \lim_{x\to\infty} \left(1+\frac1x\right)^x \qquad\qquad 2. \lim_{x\to0^+} x^x\text{.}$

Solution

Our brief revisit of limits will be rewarded in the next section where we consider improper integration. So far, we have only considered definite integrals where the bounds are finite numbers, such as $\ds \int_0^1 f(x)\ dx\text{.}$ Improper integration considers integrals where one, or both, of the bounds are “infinity.” Such integrals have many uses and applications, in addition to generating ideas that are enlightening.

# Subsection6.7.3Exercises

Terms and Concepts

In the following exercises, evaluate the given limit.