# Evaluate

Question:

Evaluate.

$\lim _{x \rightarrow 0}\left(\frac{e^{b x}-e^{a x}}{x}\right), 0 Solution: To evaluate:$\lim _{x \rightarrow 0} \frac{e^{b x}-e^{a 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{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$As$\mathrm{X} \rightarrow 0$, we have$\lim _{x \rightarrow 0} \frac{e^{b x}-e^{a x}}{x}=\frac{0}{0}$This represents an indeterminate form. Thus applying L'Hospital's rule, we get$\lim _{x \rightarrow 0} \frac{e^{b x}-e^{a x}}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(e^{b x}-e^{a x}\right)}{\frac{d}{d x}(x)}\lim _{x \rightarrow 0} \frac{e^{b x}-e^{a x}}{x}=\lim _{x \rightarrow 0} \frac{b e^{b x}-a e^{a x}}{1}\lim _{x \rightarrow 0} \frac{e^{b x}-e^{a x}}{x}=b-a$Thus, the value of$\lim _{x \rightarrow 0} \frac{e^{b x}-e^{a x}}{x}\$ is b-a.