Question

If $\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$ and $a>0$, then find the value of $a$.

Answer 1

Given: $\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$ & $a>0$

$-1=\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}$ (of the form $\frac{0}{0}$)

$-1=\underset{x\to a}{lim}\frac{a^x.\left(\ln a\right)-a{x}^{a-1}}{x^x(1+\ln x)-0}[\frac{d}{dx}.\left(x^x\right)=x^x(1+\ln x)]$ [L-Hospital Rule]

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

$-1=\frac{(\ln a-1)}{\ln a+1}$

$-\ln a-1=\ln a-1$

$\Rightarrow 2\ln a=0$

$\Rightarrow a=1$