$\underset{x\to a}{lim} \frac{x^a-a^x}{x^x-a^a}$ is equal to

Solution:

We have,

$\underset{x\to a}{lim} \frac{x^a-a^x}{x^x-a^a}$                   $\left[\frac{0}{0} form\right]$

$=\underset{x\to a}{lim} \frac{a{x}^{a-1}-a^x\log a}{x^x(1+\log x)-0}$                       [Using L' Hospital's Rule]

$=\frac{a^a-a^a\log a}{a^a(1+\log a)}=\frac{1-\log a}{1+\log a}=\frac{{\log }_e(e/a)}{{\log }_e\left(ae\right)}$