Part A:
Let $a_n=\gamma ^{s \over n}$ for some $\gamma>1$, some $s>0$, and all positive integers $n$. Show that:
$$\lim_{n\to \infty} a_n = 1$$
I've got this part, super easy.
Part B:
Fix some s>0. Determine (with proof!)
$$\lim_{n\to \infty} {\gamma^{s/n}-1\over\gamma^{1/n}-1}$$
Here's where I'm having problems! It's clear that using L'Hopitial's Rule will give me the answer I want, but the Theorem we are given says three things:
1) $f(x)$ and $g(x)$ must be continuous on $[a,b]$
2) $f(x)$ and $g(x)$ must be differentiable on $(a,b)$
3) For some $c \in [a,b], f(c)=g(c)=0$
My problem is two fold:
- The first is that I have a function that is defined across an open interval and my Theorem applies to closed intervals.
- The second is that, currently, I do not have a $c$ where $f(c)=g(c)=0$. I contemplated doing a substitution where $x=\gamma^{1\over n}$, but that requires me to change the limit from $n \to \infty$ to $x\to ?$. I don't know what I can change the limit to. Making $x \to 1$ is the obvious answer, but how do I justify it?
