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

Hi I have a question regarding of limits to infinity please help! Thank You!

The question states the user to find the following limit:

$ \lim_{n\to\infty} n^2 ({\sqrt[n]{x}-\sqrt[n+1]{x}}) $

Where $(x > 0) $

Thank You!!!

share|cite|improve this question
The limit is equivalent to $\lim_{m \to 0} \frac{x^{1/m} - x^{m/m+1}}{m^2}$. Use L'Hospital's rule twice to get the answer. – Sandeep Thilakan Mar 11 '14 at 18:56
up vote 3 down vote accepted

Rewrite it as $$ \lim_{n\to\infty}(n^2+n)\left(x^{1/(n^2+n)}-1\right)\lim_{n\to\infty}\left(\frac{n}{n+1}x^{1/(n+1)}\right) $$ and use the limit $$ \lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x) $$ which is the inverse of $$ \lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x $$

share|cite|improve this answer

Using Taylor series: $$n^2({\sqrt[n]{x}-\sqrt[n+1]{x}})=n^2\left(\exp\left(\frac1n\log x\right)-\exp\left(\frac1{n+1}\log x\right)\right)\sim_\infty n^2\log x\left(\frac1{n}-\frac1{n+1}\right)\sim_\infty \log x$$

share|cite|improve this answer
so would the final answer be 0 or infinity? – AskingQnsPro Mar 11 '14 at 18:42
No the limit of your sequence is $\log x$. – user63181 Mar 11 '14 at 18:43

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.