Question

If $\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$, then the value of $a$ is

• A. $1$
• B. $0$
• C. $e$
• D. none of these

Answer 1

The correct option is A $1$

We have,
$\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$
$\Rightarrow {lim}_{x\to a}\frac{a^{x}loga-a{x}^{a-1}}{x^x(1+logx)}=-1$ [Using L' Hospital's Rule]
$\Rightarrow \frac{a^{a}lo{g}_{e}a-a\cdot a^{a-1}}{a^a(1+lo{g}_{e}a)}=-1$
$\Rightarrow \frac{lo{g}_{e}a-1}{lo{g}_{e}a+1}=-1\Rightarrow lo{g}_{e}a=0\Rightarrow a=1$
Hence, option 'A' is correct.