Existence of the following limit

Question

I am required to calculate the following:

$\displaystyle{\lim_{x \to a}} \frac{x^a-a^x}{a^x-a^a}$, where $a>0$, $a\neq1$

My first thought would be l'Hôpital but seeing as though it has nothing to do with differentiability, we cannot use it, and so I am pretty stumped. Any help would be appreciated.

Answer 1

we have $$\lim_{x\to a}\frac{x^a-a^x}{a^x-a^a}=\lim_{x\to a}\frac{ax^{a-1}-a^x\ln(a)}{a^x\ln(a)}=-1+\frac{1}{\log(a)}$$

Answer 2

HINT

Note that

$$ \frac{x^a-a^x}{a^x-a^a}= \frac{x-a}{a^x-a^a} \frac{x^a-a^x}{x-a}=\frac{x-a}{a^x-a^a} \left(\frac{x^a-a^a+a^a-a^x}{x-a}\right)=\frac{x-a}{a^x-a^a} \left( \frac{x^a-a^a}{x-a} -\frac{a^x-a^a}{x-a} \right)$$

then use the definition of limit for $x^a$ and $a^x$ for $x\to a$.