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.
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.
Case 1: If $a>1$ then from $\exp ^{\log u(x)}=u(x)$ we deduce
$$
\begin{aligned}
\left(\frac{1}{x} \frac{a^{x}-1}{a-1}\right)^{\frac{1}{x}} &=\exp \left(\frac{\log \left(\frac{1}{x} \frac{a^{x}-1}{a-1}\right)}{x}\right) \\
&=\exp \left(\frac{\log \left(\frac{1}{x}\right)+\log \left(\frac{a^{x}-1}{a-1}\right)}{x}\right) \\
&=\exp \left(\frac{-\log (x)+\log \left(a^{x}-1\right)-\log (a-1)}{x}\right) \\
&=\exp \left(\frac{-\log (x)+\log \left[a^{x}\left(1-a^{-x}\right)\right]-\log (a-1)}{x}\right) \\
&=\exp \left(\frac{-\log (x)+\log \left(1-a^{-x}\right)-\log (a-1)}{x}+\log (a)\right)
\end{aligned}
$$
Sending $x \rightarrow \infty$ will make the middle terms vanish since
$$
\lim _{x \rightarrow \infty} \frac{\log (x)}{x}=0
$$
Therefore
$$
\lim _{x \rightarrow+\infty}\left(\frac{1}{x} \frac{a^{x}-1}{a-1}\right)^{\frac{1}{x}}=\exp (\log (a))=a
$$
Case 2: If $0<a<1$ then we first write
$$
\frac{a^{x}-1}{a-1}=\frac{1-a^{x}}{1-a}
$$
where
$$
\begin{aligned}
\log \left(\frac{1}{x} \frac{1-a^{x}}{1-a}\right) &=\log \left(\frac{1}{x}\right)+\log \left(\frac{1-a^{x}}{1-a}\right) \\
&=-\log (x)+\log \left(1-a^{x}\right)-\log (1-a)
\end{aligned}
$$
and
$$
\lim _{x \rightarrow \infty} \frac{-\log (x)+\log \left(1-a^{x}\right)-\log (1-a)}{x}=0
$$
thus
$$
\lim _{x \rightarrow+\infty}\left(\frac{1}{x} \frac{1-a^{x}}{1-a}\right)^{\frac{1}{x}}=\exp (0)=1
$$