# Find the Limit (n to infinity)

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

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 $$\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x$$
Using Taylor series: $$n^2({\sqrt[n]{x}-\sqrt[n+1]{x}})=n^2\left(\exp\left(\frac1n\log x\right)-\exp\left(\frac1{n+1}\log x\right)\right)\sim_\infty n^2\log x\left(\frac1{n}-\frac1{n+1}\right)\sim_\infty \log x$$
No the limit of your sequence is $\log x$. –  Sami Ben Romdhane Mar 11 '14 at 18:43