Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

i cannot figure out a way to find this limit. $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$$

Its undeterminate form $0\cdot\infty$ so i tried using $$\displaystyle\lim_{n\to \infty}\frac{a^{X_n}-1}{X_n}=\ln{a}$$ this leads to $0\cdot\infty$ again. I then tried to transform it in $\frac{0}{0}$ $$\displaystyle\lim_{n\to \infty} \frac{\sqrt[n]{2}-\sqrt[n+1]{2}}{\frac{1}{n^2}}$$ here i dont know what i could do to find the limit.

share|cite|improve this question
Are you allowed to use Hopital's rule? I guess, you are not. – Babak S. Jan 27 '13 at 10:45
Hint: $e^x=1+x+x^2/2+O(x^3)$ as $x\to0$. – Hanul Jeon Jan 27 '13 at 10:45
I am not allowed to use Hopital's rule and i do not know big O notation. – phi Jan 27 '13 at 10:46
up vote 10 down vote accepted


$$2^{\frac{1}{n}} = e^{\frac{\log{2}}{n}}$$

so that

$$\begin{align} \lim_{n \rightarrow \infty} n^2 \left (2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right ) &= \lim_{n \rightarrow \infty} n^2 \left (e^{\frac{\log{2}}{n}} - e^{\frac{\log{2}}{n+1}} \right ) \\ &= \lim_{n \rightarrow \infty} n^2 \log{2} \left ( \frac{1}{n} - \frac{1}{n+1}\right ) \\ &= \lim_{n \rightarrow \infty} \frac{n \log{2}}{n+1} \\ &= \log{2} \\ \end{align} $$

The second step relies on the fact that

$$e^x = 1+ x + O(x^2)$$

as $x \rightarrow 0$.

share|cite|improve this answer
You actually have to be a little more careful here: while it's true that the $O(x^2)$ terms cancel when you expand out $e^x$ for $x=\frac{\log 2}{n}$ and $x=\frac{\log 2}{n+1}$, that's not necessarily self-evident on the face of it, and of course because we're multiplying by $n^2$, if those second-order terms didn't cancel then they'd contribute a finite amount to the solution. Better to expand out to order 3. – Steven Stadnicki Jan 28 '13 at 3:24
@StevenStadnicki: You make a good point, although as you know the $k$th-order term provides a contribution of $O(n^{1-k})$ to the limit piece, which of course goes to zero in this limit when $k>1$. I could have been more explicit about that. – Ron Gordon Jan 28 '13 at 10:04

By the mean value theorem $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})=\displaystyle\lim_{n\to \infty} \left(n^2 \times\frac{2^{1/c_{n}}\ln 2}{c_n^2}\right)=\ln2$$ where $n<c_n<n+1$


share|cite|improve this answer
this is an outstanding solution – user29743 Jan 28 '13 at 2:33
@countinghaus To be fair, L'Hôpital is just a clever application of the intermediate value theorem too. – Pedro Tamaroff Feb 27 '13 at 2:32

We have:

$ \begin{align*} \lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2}) & = \lim_{n\to \infty} n^2\left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n}-\frac{1}{n+1}} - 1 \right) \\ & = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right) \\ & = \lim_{n\to \infty} \cfrac{n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right)}{\cfrac{1}{n^2+n}(n^2+n)} \\ & = \lim_{n\to \infty} \dfrac{n^2}{n^2+n} \cdot \lim_{n\to \infty} 2^{\frac{1}{n+1}} \cdot \lim_{n\to \infty} \cfrac{2^{\frac{1}{n^2+n}} - 1}{\cfrac{1}{n^2+n}} \\ & = 1 \cdot 1 \cdot \ln 2 \\ & = \boxed{\ln 2}. \end{align*} $

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.