<s>[INST] 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.
 [/INST] 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
$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{1-a^x}{1-a} \right)^\frac{1}{x}=\exp\bigl(0)=1.$$
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