L'Hopital's Rule

Two linear functions are given in Fig. 1 , and we need to find $\lim \limits_{x \rightarrow 5} \frac{f(x)}{g(x)}$. Unfortunately, $\lim \limits_{x \rightarrow 5} f(x)=0$ and $\lim \limits_{x \rightarrow 5} g(x)=0$ so we cannot apply the Main Limit Theorem. However, we know $\mathrm{f}$ and $\mathrm{g}$ are linear, we can calculate their slopes from Fig. 1 , and we know that they both go through the point $(5,0)$ so we can find their equations: $f(x)=-2(x-5)$ and $g(x)=3(x-5)$.

Fig. 1

Now the limit is easier: $\lim _{x \rightarrow 5} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 5} \frac{-2(x-5)}{3(x-5)}=\frac{-2}{3}=\frac{\text { slope of } \mathrm{f}}{\text { slope of } \mathrm{g}}$.

In fact, this pattern works for any two linear functions:

The really powerful result, discovered by John Bernoulli and named for the Marquis de 1'Hô pital who published it in his calculus book, is that the same pattern is true for differentiable functions even if they are not linear.

Idea for a proof: Even though $\mathrm{f}$ and $\mathrm{g}$ may not be linear functions, they are differentiable so at the point $\mathrm{x}=\mathrm{a}$ they are "almost linear" in the sense that they are well approximated by their tangent lines at that point (Fig. 2):

since $f(a)=g(a)=0$ $f(x) \approx f(a)+f^{\prime}(a)(x-a)=f^{\prime}(a)(x-a)$ and $g(x) \approx g(a)+g^{\prime}(a)(x-a)=g^{\prime}(a)(x-a)$

Fig. 2

(Unfortunately, we have ignored a couple subtle difficulties such as $\mathrm{g}(\mathrm{x})$ or $\mathrm{g}^{\prime}(\mathrm{x})$ possibly being $0$ when $\mathrm{x}$ is close to $a$. A proof of $1^{\prime}$ Hô pital's Rule is difficult and is not included.)

Example 1: Use 1'Hô pital's Rule to determine $\lim \limits_{x \rightarrow 0} \frac{x^{2}+\sin (5 x)}{3 x}$ and $\lim \limits_{x \rightarrow 1} \frac{\ln (x)}{e^{x}-e}$.

Solution: (a) We could evaluate this limit without 1'Hô pital's Rule but let's use it. We can match the pattern of 1 'Hô pital's Rule by letting $a=0, f(x)=x^{2}+\sin (5 x)$ and $g(x)=3 x$. Then $f(0)=0, g(0) = 0$, and $\mathrm{f}$ and $\mathrm{g}$ are differentiable with $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+5 \cos (5 \mathrm{x})$ and $\mathrm{g}^{\prime}(\mathrm{x})=3$ so

$\lim _{x \rightarrow 0} \frac{x^{2}+\sin (5 x)}{3 x}=\frac{\mathrm{f}^{\prime}(0)}{\mathrm{g}^{\prime}(0)}=\frac{2 \cdot 0+5 \cos (5 \cdot \mathbf{0})}{3}=\frac{5}{3}$

(b) Let $a=1, f(x)=\ln (x)$ and $g(x)=e^{x}- e$. Then $f(1)=0, g(1)=0, f$ and $g$ are differentiable for $x$ near $1(x \neq 0)$, and $f^{\prime}(x)=1 / x$ and $g^{\prime}(x)=e^{x}$. Then

$\lim _{x \rightarrow 1} \frac{\ln (x)}{e^{x}-e}=\frac{\mathrm{f}^{\prime}(\mathbf{1})}{\mathrm{g}^{\prime}(\mathbf{1})}=\frac{1 / \mathbf{1}}{\mathrm{e}^{\mathbf{1}}}=\frac{1}{\mathrm{e}}$

Practice 1: Use 1'Hô pital's Rule to find $\lim \limits_{x \rightarrow 0} \frac{1-\cos (5 x)}{3 x}$ and $\lim \limits_{x \rightarrow 2} \frac{x^{2}+x-6}{x^{2}+2 x-8}$.