L'Hôpital's rule

In various instances in this chapter, we discussed that certain limits yield indeterminate forms; these can be $0 / 0$ or $\infty / \infty, \infty-\infty,(0)( \pm \infty), 0^{0}, \infty^{0}$ and others. When we encountered an indeterminate form in the examples above, we tried tricks like simplifying (by factorisation) or introducing a new variable. Series expansions which are discussed in Chapter 6 are also very useful in these cases. However, L'Hôpital's rule is a quick method for computing limits which have an indeterminate form of $0 / 0$ or $\infty / \infty$ . While L'Hôpital's rule is given in Definition 2.3 in this section, note that its application requires knowledge of differentiation which is the subject of Chapter 3.

${ }^{5}$ Let $f_{n}(x)=(1+x / n)^{n}$ and differentiate wrt $x$ . With $f(x)$ and $f^{\prime}(x)$ defined as $f(x)=\lim _{n \rightarrow \infty} f_{n}$ and $f^{\prime}(x)=\lim _{n \rightarrow \infty} f_{n}^{\prime}$ , we have $f^{\prime}(x)=f(x)$ where prime denotes differentiation; this is a first order ODE with solution $f(x)=e^{x}$ satisfying $f(0)=1$ . See Chapter 12 for details.

Definition of L'Hôpital's rule

Suppose that we have one of the two following cases

\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\infty}{\infty}

where $a$ is a real number (can also be $\pm \infty$ ). Then, we have

\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}

Example 2.5 Evaluate the limit given by

\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1} .

Solution Note that the limit is of the form 0/0 and we computed it in Subsec. 2.1.1 by factoring the numerator. Now, using L'Hôpital's rule, we have

\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}=\lim _{x \rightarrow 1} \frac{2 x}{1}=2

Note that while L'Hôpital's rule as defined in Definition 2.3 only works on quotients, we can use it to compute limits of the indeterminate form $(0)( \pm \infty)$ by rewriting a product of functions $f(x) g(x)$ as quotients

f(x) g(x)=\frac{f(x)}{1 / g(x)}, \quad \text { or } \quad f(x) g(x)=\frac{g(x)}{1 / f(x)} .

We use this result in Example 2.6.

Example 2.6 Evaluate the following limit

\lim _{x \rightarrow-\infty} x e^{x}

Solution The limit is of the form $(-\infty)(0)$ . Rewriting the product $x e^{x}$ as $x / e^{-x}$ , we have

\lim _{x \rightarrow-\infty} \frac{x}{e^{-x}}=\lim _{x \rightarrow-\infty} \frac{1}{-e^{-x}}=0 .

Note that choosing $e^{x}$ as the numerator and $1 / x$ as the denominator does not help us evaluate the limit. While sometimes either quotient works, this is not always the case.

Examples Continuity