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I found the limit $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ by first defining $f(x)=\sqrt[x]{b^{2^{-x}}-1}$ above $R$ and then finding the limit of $ln(f)$ (to cancel the nth root). This worked (the result is $1/2$), but I ended up having to find the derivative of rather complex functions when I used L'hopital (twice). My worry is that if I have to solve something like this in a test I'll easily make a technical error. I was wondering if there is a simpler way to find this limit?

I know most basic techniques of finding limits in $R$ and a bit (Stoltz, Cantor's lemma, ...) about finding limits of sequences.

Thank you for your help!

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    I don't have the time to see if the argument goes anywhere, but you can use the fact that $a^n-b^n=(a-b)(a^{n-1}b+a^{n-2}b^2+\cdots+ab^{n-1})$, letting $a=\sqrt[n]{b^{2^{-n}}-1}$ and $b=\frac{1}{\sqrt[n]{2^n}}$.2011-11-07
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    IMHO - l`Hopital is one of the simplest methods. It saves you proofs of the form "for each epsilon there exists N such that for every n>N ...". Those usually involve more complex technics2011-11-07
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    +1 for explaining your reasoning. We've had many questions recently that ask, inexplicably, for help with "_(doing something or other)_ without using _(simple standard first-choice technique)_".2011-11-07
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    Tests seldom have questions that are technically very demanding. Finding a clever solution ordinarily takes more time than applying standard techniques in more or less standard ways. The problem with a truly good solution like the one by robjohn is that you may be downgraded for not filling in details. The suggestion by Eric Naslund helps. Let $y=2^{-x}$ and you are looking at the limit as $y$ approaches $0$ of $(-\ln 2)\frac{\ln(b^y-1)}{\ln y}$, still two applications of L'Hospital's Rule, but less risky differentiation.2011-11-07

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