# Limit of $(2-\sqrt[n]{2})^n$?

As put forth in the title, the question is to find:

$$\lim_{n\to\infty}(2-\sqrt[n]{2})^n$$

Wolfram Alpha leads me to believe that the correct answer is $\dfrac{1}{2}$.

Thank you!

edit: $n$ is going to infinity. I tried taking the natural log, and did this did not help me. I also attempted to rewrite the term as $$\left[(1+(1-\sqrt[n]{2}))^{\frac{1}{1-\sqrt[n]{2}}}\right]^{n(1-\sqrt[n]{2})},$$ but also got nowhere.

• A very important detail which is messing in your question!!! Where is $n$ going? $n \rightarrow ?$. Secondly, why $Wolfram$?... just asking :) – Sergio Sarmiento Jun 4 '14 at 13:43
• Add your research effort to the question. – Apurv Jun 4 '14 at 13:44
• I tried Wolfram because I thought I got the answer to be 1 at some point and did not believe it. Plugging in high values of $n$ led me to 1/2. I discovered my mistake quickly after realizing I "must" be wrong. – user88849 Jun 4 '14 at 13:58
• The form of the problem suggests the definition of $e$. See robjohn's comment for the development. – Snowbody Jun 4 '14 at 18:35

We have

$$\log(2-\sqrt[n]2)=\log(1+(1-\sqrt[n]2))$$ and $$1-\sqrt[n]2=1-\exp\left(\frac1n\log2\right)\sim_\infty-\frac{\log2}n$$ hence we see that $$\log(2-\sqrt[n]2)\sim_\infty -\frac{\log2}n$$ so finally $$\lim_{n\to\infty}(2-\sqrt[n]{2})^n=\lim_{n\to\infty}\exp\left(-\frac{n\log2}n\right)=\frac12$$

This solution uses L'Hopital's rule. $$\lim_{n\to \infty}(2-\sqrt[n]{2})^n=\lim_{n\to \infty}e^{n\ln (2-\sqrt[n]{2})}=e^{\lim_{n\to \infty}\frac{\ln(2-\sqrt[n]{2})}{1/n}}$$

$$\lim_{n\to \infty}\frac{\ln(2-\sqrt[n]{2})}{1/n}=\lim_{n\to \infty}\frac{\frac{1}{2-\sqrt[n]{2}}(\frac{\ln 2}{n^2}\sqrt[n]{2})}{-1/n^2}=\lim_{n\to \infty}\frac{1}{2-\sqrt[n]{2}}\lim_{n\to \infty}-(\ln 2 )\sqrt[n]{2}=-\ln 2$$

Now plug in to get $e^{-\ln 2}=1/2$

Note that \begin{align} \left[1+\frac{n\left(2^{1/n}-1\right)}n\right]^n &=\left[2^{1/n}\right]^n\\ &=2\tag{1} \end{align} Since $$\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\tag{2}$$ converges uniformly on compact subsets of $\mathbb{R}$ and $e^x$ is strictly increasing, $(1)$ and $(2)$ imply $$\lim_{n\to\infty}n\left(2^{1/n}-1\right)=\log(2)\tag{3}$$ Again, because $(2)$ converges uniformly on compact subsets of $\mathbb{R}$, \begin{align} \lim_{n\to\infty}\left(2-2^{1/n}\right)^n &=\lim_{n\to\infty}\left(1+1-2^{1/n}\right)^n\\ &=\lim_{n\to\infty}\left[1+\frac{n\left(1-2^{1/n}\right)}n\right]^n\\ &=\lim_{n\to\infty}\left[1+\frac{-\log(2)}n\right]^n\\[9pt] &=e^{-\log(2)}\\[12pt] &=\frac12\tag{4} \end{align}

• Oops, was meant to be a comment on the question pointing out your answer, which seems to me the best way to get the answer – Snowbody Jun 4 '14 at 18:36

Hint: if $a_n>0$ for all $n$ and $\lim_{n\to\infty} a_n = a \neq 0,$ then $\lim_{n\to\infty} \ln(a_n) = \ln a.$

• $\;a_n, a>0\;$ ........... – DonAntonio Jun 4 '14 at 13:53