Question

If $\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$ find 'a'

• A. $a=1$
• B. $a=1/e$
• C. $a=e$
• D. None of these.

Answer 1

The correct option is A $a=1$

$\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$
Since, it is of the form $\frac{0}{0}$ , so applying L-hopital's rule
$\underset{x\to a}{lim}\frac{a^x{\log }_{e}a-a{x}^{a-1}}{x^x(1+{\log }_{e}x)}=-1$
$\Rightarrow \frac{a^a{\log }_{e}a-a{a}^{a-1}}{a^a(1+{\log }_{e}a)}=-1$
$\Rightarrow {\log }_{e}a-1=-1-{\log }_{e}a$
$\Rightarrow {\log }_{e}a=0$
$\Rightarrow a=1$