Evaluate

Solution:

To evaluate:

$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$

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

$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}=\frac{0}{0}$

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

$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(a^{x}-b^{x}\right)}{\frac{d}{d x}(x)}$

$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}=\lim _{x \rightarrow 0} \frac{a^{x} \ln a-b^{x} \ln b}{1}$

$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}=\ln a-\ln b$

$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}=\ln \frac{a}{b}$

Thus, the value of $\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$ is $\ln \frac{a}{b}$.