The Limit of $x\left(\sqrt[x]{a}-1\right)$ as $x\to\infty$. How to find the limit of:
$$ \lim_{x\to \infty }x\left(\sqrt[x]{a}-1\right)$$
Without using L'hôspital rule.
I've tried to bound the term and use the squeeze theorem but I couldn't find the right upper bound. I've also tried to convert $a^\frac{1}{x}$ to $e^{\frac{1}{x}\ln{a}}$ but it didn't help me.
Whats the right way to evaluate that limit?

Setting $\displaystyle \frac1x=h$
$$\lim_{n\to\infty}x(\sqrt[x]a-1)=\lim_{h\to0}\frac{a^h-1}h=\ln a\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}=\cdots$$
Use Proof of $ f(x) = (e^x-1)/x = 1 \text{ as } x\to 0$ using epsilon-delta definition of a limit
