Finding $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ without L'hopital

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!

Solution 1:

Use $\left(b^{2^{-n}}-1\right) 2^n \sum_{k=0}^{2^n-1} b^{k 2^{-n}} 2^{-n} = b - 1$. Notice that, by the definition of Riemann integral, $\lim_{n \to \infty} \sum_{k=0}^{2^n-1} b^{k 2^{-n}} 2^{-n} = \int_0^1 b^x \mathrm{d} x = \frac{b-1}{\log b}$.

Hence the result: $$ \lim_{n \to \infty} \sqrt[n]{b^{2^{-n}}-1} = \frac{1}{2} \lim_{n \to \infty} \sqrt[n]{ \log b } = \frac{1}{2} $$

Solution 2:

For $x$ near $0$, the Taylor series for $e^x-1$ gives $$ e^x-1=x+O(x^2)\tag{1} $$ so $$ \begin{align} b^{2^{-n}}-1 &=e^{\log(b)\;2^{-n}}-1\\ &=\log(b)\;2^{-n}+O(4^{-n})\\ &=\log(b)\;2^{-n}(1+O(2^{-n}))\tag{2} \end{align} $$ It is fairly easy to show that the limit of $n^{th}$ root of $(2)$ is $2^{-1}=\frac{1}{2}$

Solution 3:

Taking the logarithm is a very good way to start. Then we are looking at $$\lim_{n\rightarrow \infty } \frac{1}{n}\log\left(b^{2^{-n}}-1\right).$$ Do a variable change, and let $x=2^{-n}$ so that this is $$\lim_{x\rightarrow 0} -\frac{\log 2}{\log x} \log\left( b^x -1\right).$$ As $b^x=e^{x\log b} =1 +x\log b+O(x^2)$, we see that this is

$$\lim_{x\rightarrow 0} -\frac{\log 2}{\log x}(\log(x\log b)+\log(1+O(x)))=\lim_{x\rightarrow 0}-\log 2+O\left(\frac{1}{\log x}+x\right)=-\log 2.$$

Hence the original limit is $e^{-\log 2}=\frac{1}{2}$.