Please help to evaluate this limit
$$
\lim _{x \rightarrow \infty}\left(\frac{1}{x} \frac{a^{x}-1}{a-1}\right)^{\frac{1}{x}}
$$
where $0 \leq a$ and $a \neq 1$
I tried to logarithm from both sides, and apply taylor series but so far without success.
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Take the (natural) logarithm. For $a>1,$ we have
$$
\lim _{x \rightarrow \infty} \frac{-\log x+\log \left(a^{x}-1\right)-\log (a-1)}{x}=\lim _{x \rightarrow \infty} \frac{\log \left(a^{x}-1\right)}{x}=\lim _{x \rightarrow \infty} \frac{a^{x} \log a}{a^{x}-1}=\log a
$$
with a simple application of l’Hôpital.
For $0<a<1,$ you get
$$
\lim _{x \rightarrow \infty} \frac{-\log x+\log \left(1-a^{x}\right)-\log (1-a)}{x}=0
$$
which needs nothing special.
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