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