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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!!!

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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

2 Answers 2

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 $$ \lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x $$

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

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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

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