Examine the limit when $x$ approaches infinity Examine:
$$\lim_{x \to \infty} \left(\frac{1}{x} \times \frac{a^x -1}{a-1}\right)^\frac{1}{x}$$
Tips of hints to point me in the right direction?
 A: Since\begin{align}\left(\frac1x\times\frac{a^x-1}{a-1}\right)^\frac1x&=\exp\left(\frac{\log\left(\frac1x\times\frac{a^x-1}{a-1}\right)}x\right)\\&=\exp\left(\frac{\log(a^x-1)-\log(x)-\log(a-1)}x\right)\\&=\exp\left(\log(a)+\frac{\log(1-a^{-x})-\log x-\log(a-1)}x\right)\end{align}and since $\lim_{x\to+\infty}\frac{\log x}x=0$,$$\lim_{x\to+\infty}\left(\frac1x\times\frac{a^x-1}{a-1}\right)^\frac1x=\exp\bigl(\log(a)\bigr)=a.$$
A: $$\ln\left(\left(\frac{a^x-1}{(a-1)x} \right)^{1/x}\right)=\frac{\ln\lvert a^x-1\rvert-\ln\lvert a-1\rvert-\ln x}{x}$$
Now, for the asymptotic behaviour of $\ln\lvert a^x-1\rvert$ there are two cases...
A: First, the expression $\left({1\over x}\times{a^x-1\over a-1}\right)^{1\over x}$ only makes (real) sense if $a\ge0$ and $a\not=1$.  Note also that $a^x\gt1$ if and only if $a\gt1$ (for $x\gt0$), so we can rewrite the expression with absolute values:
$$\left({1\over x}\times{|a^x-1|\over |a-1|}\right)^{1\over x}$$
Now $\lim_{x\to\infty}(1/x)^{1/x}=\lim_{u\to0^+}u^u=1$ is a standard limit, as is $\lim_{x\to\infty}|a-1|^{1/x}=1$ when $a\not=0$. Thus
$$\lim_{x\to\infty}\left({1\over x}\times{|a^x-1|\over |a-1|}\right)^{1\over x}=\lim_{x\to\infty}|a^x-1|^{1/x}$$
Here's where it gets a little tricky: The cases $0\le a\lt1$ and $a\gt1$ must be treated separately.
For $0\le a\lt1$, we have $a^x\to0$ as $x\to\infty$, so $|a^x-1|=1-a^x\to1$, from which it follows that $|a^x-1|^{1/x}\to1$.  
For $a\gt1$, we can write $|a^x-1|=a^x(1-r^x)$ with $r=1/a$ and thus $|a^x-1|^{1/x}=a|1-r^x|^{1/x}\to a$ since $a\gt1$ implies $0\le r\lt1$. Putting everything together, we have
$$\lim_{x\to\infty}\left({1\over x}\times{|a^x-1|\over |a-1|}\right)^{1\over x}=\lim_{x\to\infty}|a^x-1|^{1/x}
=\begin{cases}
1\quad\text{if }0\le a\lt1\\
a\quad\text{if }a\gt1
\end{cases}$$
