Evaluate
$\lim _{x \rightarrow 0}\left(\frac{e^{2+x}-e^{2}}{x}\right)$
To evaluate:
$\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}$
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
$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$
As $\mathrm{x} \rightarrow 0$, 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
$\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(e^{2+x}-e^{2}\right)}{\frac{d}{d x}(x)}$
$\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}=\lim _{x \rightarrow 0} \frac{e^{2+x}}{1}$
$\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}=e^{2}$
Thus, the value of $\lim _{x \rightarrow 0} \frac{e^{2+x}-e^{2}}{x}$ is $e^{2}$.
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