TITLE: Finding $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ without L'hopital
QUESTION [16 upvotes]: 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!

REPLY [3 votes]: If $(c_n)$ is a sequence of positive numbers such that $\lim\limits_{n\to\infty}\frac{c_{n+1}}{c_n}$ exists, then $\lim\limits_{n\to\infty}\sqrt[n]{c_n}=\lim\limits_{n\to\infty}\frac{c_{n+1}}{c_n}$.  See for example Theorem 3.37 of Rudin's Principles of mathematical analysis, 3rd Ed.  
If we let $c_n=b^{2^{-n}}-1$, then $$\frac{c_{n+1}}{c_n}=\frac{\sqrt{b^{2^{-n}}}-1}{b^{2^{-n}}-1}=\frac{1}{\sqrt{b^{2^{-n}}}+1}\to\frac{1}{2}.$$