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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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    @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

1 Answers 1

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Using the following limit $\lim\limits_{x\to 0}{\dfrac{a^x-1}{x}}=\ln{a}$ we have $\lim\limits_{n\to \infty} {\dfrac{\gamma^{\tfrac{s}{n}}-1}{\gamma^{\tfrac{1}{n}}-1}}=\lim\limits_{n\to \infty} \left({\dfrac{\gamma^{\tfrac{s}{n}}-1}{\frac{s}{n}}}\cdot {\dfrac{{\frac{s}{n}}}{\gamma^{\tfrac{1}{n}}-1}}\right)=s\lim\limits_{n\to \infty} \left({\dfrac{\gamma^{\tfrac{s}{n}}-1}{\frac{s}{n}}}\cdot {\dfrac{{\frac{1}{n}}}{\gamma^{\tfrac{1}{n}}-1}}\right)=\\ =s\lim\limits_{n\to \infty} {\dfrac{\gamma^{\tfrac{s}{n}}-1}{\frac{s}{n}}} \cdot \lim\limits_{n\to \infty} {\dfrac{{\frac{1}{n}}}{\gamma^{\tfrac{1}{n}}-1}}= s\dfrac{\ln{\gamma}}{\ln{\gamma}}=s.$