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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?
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    The hypotheses in L'Hopital's Rule apply to the variable for which the limit is being taken.2012-12-16
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    $f(\gamma)=\gamma^{\frac{s}{n}}-1$, $g(\gamma)=\gamma^{\frac{1}{n}}-1$ and $c=1$ ;)2012-12-16
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    @Jonathan: Please don't delete the body of your question, even after you've had it answered. That just confuses other users of the site.2012-12-17

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