Limit with a parameter I was trying to solve this limit while preparing for my finals:
$$\lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x-1}{a-1}\right)^{\frac{1}{x}}$$
Here are my steps:
$$g=\lim_{x\to\infty} \left(\frac{1}{x}\cdot\frac{a^x-1}{a-1}\right)^{\frac{1}{x}}$$
$$\ln g = \lim_{x\to\infty}\frac{\ln\left(\frac{1}{x}\cdot\frac{a^{x}-1}{a-1}\right)}{x}$$
Using L'Hospital:
$$\ln g = \lim_{x\to\infty}\frac{a^{x}\cdot\ln a\left(ax-x\right)-\left(a-1\right)\cdot\left(a^{x}-1\right)}{\left(a^{x}-1\right)\left(ax-x\right)}$$
Here is where I got stuck. Using another L'Hospital would result in an algebraic mess that would not get me anywhere. I will appreciate any push in the correct direction. Thank you.
 A: Notice that, for all $\alpha>0$, $\lim_{x\to\infty}(\alpha x)^{1/x}=1$.
If $0\le a<1$, then $\lim_{x\to\infty}\left(1-a^x\right)^{1/x}=\left[(1-0)^0\right]=1$ and therefore $$\lim_{x\to\infty}\left(\frac{a^x-1}{x(a-1)}\right)^{1/x}=\lim_{x\to\infty}\frac{\left(1-a^x\right)^{1/x}}{((1-a)x)^{1/x}}=1.$$
If $a>1$, then $\lim_{x\to\infty}(a^x-1)^{1/x}=\lim_{x\to\infty}a(1-(a^{-1})^x)^{1/x}=a$ by the previous remark.
Therefore $$\lim_{x\to\infty}\left(\frac{a^x-1}{x(a-1)}\right)^{1/x}=\lim_{x\to\infty}\frac{(a^x-1)^{1/x}}{(x(a-1))^{1/x}}=a.$$
A: Following your method, expand like this:
$$\frac{\ln\left(\frac{1}{x}\cdot\frac{a^x-1}{a-1}\right)}{x}=-\frac{\ln(x)}{x}+\frac{\ln\left(\frac{a^x-1}{a-1}\right)}{x}$$
Note that $$\lim_{x\to\infty}\frac{\ln(x)}{x}=0$$
If $a<0$, then this limit is not well defined. If $a\geq 1$, then $(a^x-1)/(a-1)$ is divergent as $x\to\infty$, so assuming $0\leq a<1$, we have that $$\lim_{x\to\infty}\frac{a^x-1}{a-1}=\frac{1}{1-a}$$ and since $\ln$ is continuous, we have that $$\lim_{x\to\infty}\ln\left(\frac{a^x-1}{a-1}\right)=\ln\left(\frac{1}{1-a}\right)=-\ln(1-a)$$
Then $$\lim_{x\to\infty}\frac{\ln\left(\frac{a^x-1}{a-1}\right)}{x}=0$$
Since $\ln g = 0$, then $g=1$.
