Evaluate

Solution:

To evaluate

$\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}$

Formula used:

L'Hospital's rule

Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where

$\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$

then

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

As $x \rightarrow 0$, we have

$\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}=\frac{0}{0}$

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

$\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}=\lim _{x \rightarrow 4} \frac{\frac{d}{d x}\left(e^{x}-e^{4}\right)}{\frac{d}{d x}(x-4)}$

$\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}=\lim _{x \rightarrow 4} \frac{e^{x}}{1}$

$\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}=e^{4}$

Thus, the value of $\lim _{x \rightarrow 4} \frac{e^{x}-e^{4}}{x-4}$ is $e^{4}$.