$\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

How does the n^2/3 distribute?

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- Nov 24th 2012, 07:44 PMsuperduper1Calculate the limit as n-> infinity
$\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

How does the n^2/3 distribute? - Nov 24th 2012, 08:03 PMMarkFLRe: Calculate the limit as n-> infinity
$\displaystyle n^{\frac{2}{3}}=\sqrt{n^{\frac{4}{3}}}$

- Nov 24th 2012, 08:22 PMsuperduper1Re: Calculate the limit as n-> infinity
How? Forgive my innocence.

- Nov 24th 2012, 08:30 PMjakncokeRe: Calculate the limit as n-> infinity
- Nov 24th 2012, 08:40 PMsuperduper1Re: Calculate the limit as n-> infinity
$\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

$\displaystyle \lim_{n\rightarrow \infty} \frac{\sqrt{n^\frac{7}{3}+2n^\frac{4}{3}}}{2n^2+n+ 1}$

What should I do next? - Nov 24th 2012, 08:45 PMMarkFLRe: Calculate the limit as n-> infinity
Basically, if you take a positive expression under a square root, you double the exponent. In general:

$\displaystyle a^b=\sqrt[n]{a^{bn}}$ where $\displaystyle 0\le a^b$ if $\displaystyle n$ is even and $\displaystyle b$ is odd. - Nov 24th 2012, 08:48 PMMarkFLRe: Calculate the limit as n-> infinity
- Nov 24th 2012, 09:22 PMsuperduper1Re: Calculate the limit as n-> infinity
solutions says it is 1/2. I keep getting 1/infinity

- Nov 24th 2012, 09:50 PMMarkFLRe: Calculate the limit as n-> infinity
I get zero as well, and so does W|A. Are you certain you have stated the problem correctly? The exponent on n outside the radical in the numerator would have to be 7/2 in order for the limit to be 1/2.

- Nov 24th 2012, 10:37 PMMarkFLRe: Calculate the limit as n-> infinity
Oops, sorry make that 3/2, not 7/2, which makes more sense for a type to have occured.

- Nov 25th 2012, 04:30 AMPlatoRe: Calculate the limit as n-> infinity