Find the limit $\displaystyle\lim_{x\to\infty}\left(\dfrac{a^x-1}{x(a-1)}\right)^{1/x}$.
While solving limits from my book I found this in the exercise. It says the answer is $1$ for $a<1$ and $a$ for $a>1$. I have tried different method like taking log. All I could prove that for $a>1$ the limit is greater than $1$.
First consider $a > 1$ and then we can write $$f(x) = \frac{1}{(a - 1)^{1/x}}\cdot\frac{1}{x^{1/x}}\cdot(a^{x} - 1)^{1/x}$$ Now $(a - 1) > 0$ so $(a - 1)^{1/x} \to (a - 1)^{0} = 1$ as $x \to \infty$. Also $x^{1/x} = \exp((\log x)/x) \to \exp(0) = 1$ as $x \to \infty$. The last factor can be written $a(1 - (1/a^{x}))^{1/x}$ which tends to $a(1 - 0)^{0} = a$. So finally $f(x) \to a$ as $x \to \infty$.
If $a < 1$ then we write $$f(x) = \frac{1}{(1 - a)^{1/x}}\cdot\frac{1}{x^{1/x}}\cdot(1 - a^{x})^{1/x}$$ and the first two factors tend to $1$. Also since $0 < a < 1$ we see that $a^{x} \to 0$ and hence the third factor $(1 - a^{x})^{1/x} \to (1 - 0)^{0} = 1$. So in this case $f(x) \to 1$ as $x \to \infty$.
hint: Assume $a > 1$:$\ln f(x) = \dfrac{\ln(a^x-1)-\ln x-\ln(a-1)}{x}$, and use L'hospitale rule. You can do the case $a < 1$.
