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.
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