Question

If $a>0$ and $\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$ then $a$ =

• A. $0$
• B. $1$
• C. $e$
• D. $2e$

Answer 1

The correct option is B $1$

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

LHS is of the form $\frac{0}{0}$

So, applying L-Hospital's rule
$\underset{x\to a}{lim}\frac{a^x\log a-a{x}^{a-1}}{x^x(1+\log x)}=-1$

$\Rightarrow \frac{a^a(\log a-1)}{a^a(1+\log a)}=-1$

$\Rightarrow 2a^a\log a=0$

$\Rightarrow a=1$