Evaluate this limit $ \lim_{x\to\infty}\left (\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}$ Please help to evaluate this limit
$$ \lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x},$$
where $0 \leq a$ and $a \not= 1$.
I tried to logarithm from both sides, and apply taylor series but so far without success.
 A: The limit of $\left(\frac1x\right)^{\frac1x}$ equals one (by taking logs). If $a>1,$ then $(a-1)^{\frac1x} \to 1,$ while $(a^x-1)^{\frac1x} \to a,$ so the limit equals $a.$ If $a<1,$ then the limit equals $1$ (exercise).
A: Take the (natural) logarithm. For $a>1$, we have
$$
\lim_{x\to\infty}\frac{-\log x+\log(a^x-1)-\log(a-1)}{x}=
\lim_{x\to\infty}\frac{\log(a^x-1)}{x}=\lim_{x\to\infty}\frac{a^x\log a}{a^x-1}=\log a
$$
with a simple application of l'Hôpital.
For $0<a<1$, you get
$$
\lim_{x\to\infty}\frac{-\log x+\log(1-a^x)-\log(1-a)}{x}=0
$$
which needs nothing special.
A: Case 1: If $a>1$ then from  $\text{exp}^{\large{\log{u(x)}}}=u(x)$ we deduce
\begin{align}
\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 1x\right)+\log\left(\frac{a^x - 1}{a - 1}\right)}x\right)\\&=
\exp\left(\frac{-\log(x)+\log(a^x-1)-\log(a-1)}x\right)\\&=
\exp\left(\frac{-\log(x)+\log\big[a^x\left(1-a^{-x}\right)\big]-\log(a-1)}x\right)\\&=
\exp\left(\frac{-\log (x)+\log(1-a^{-x})-\log(a-1)}x+\log(a)\right)
\end{align}
Sending $x\to \infty$ will make the middle terms vanish since
$$\lim_{x\to\infty}\frac{\log(x)}{x}=0$$
Therefore
$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}=\exp\bigl(\log(a)\bigr)=a.$$
Case 2: If $0< a <1$ then we first write
$$\frac{a^x - 1}{a - 1} = \frac{1-a^x}{1-a}$$
where
\begin{align}
\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\left({1-a}\right)
\end{align}
and
$$\lim_{x\to\infty}\frac{-\log(x)+\log\left(1- a^x\right)-\log\left({1-a}\right)}{x}=0$$
thus
$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{1-a^x}{1-a} \right)^\frac{1}{x}=\exp\bigl(0)=1.$$
