The limit as $n$ approaches infinity of $n\left(a^{1/n}-1\right)$ I need to know how to calculate this without using l'hospitals rule:
limit as $x$ approaches infinity of: $$x\left(a^{1/x}-1\right)$$
I saw that the answer is $\log(a)$, but I want to know how they got it.
The book implies that I should be able to find it by just using algebraic manipulation and substitution.

The exponential function is a convex function, hence $x<y$ gives:
$$ e^x \leq \frac{e^y-e^x}{y-x} \leq e^{y} \tag{1}$$
and assuming $a>1$ we have:
$$ e^{0}\leq \frac{e^{x\log a}-e^0}{x\log a-0}\leq e^{x\log a}\tag{2}$$
or:
$$ \log a \leq \frac{a^x-1}{x}\leq a^x \log a \tag{3}$$
hence the claim follows by squeezing.
