Evaluate

To evaluate:

$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{x}$

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 $x \rightarrow 0$, we have

$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{x}=\frac{0}{0}$

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

$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(e^{3 x}-e^{2 x}\right)}{\frac{d}{d x}(x)}$

$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{x}=\lim _{x \rightarrow 0} \frac{3 e^{3 x}-2 e^{2 x}}{1}$

$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{x}=3-2$

$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{x}=1$

Thus, the value of $\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{2 x}}{x}$ is 1 .