# Thread: Calculate the limit as n-> infinity

1. ## Calculate the limit as n-> infinity

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

How does the n^2/3 distribute?

2. ## Re: Calculate the limit as n-> infinity

$n^{\frac{2}{3}}=\sqrt{n^{\frac{4}{3}}}$

3. ## Re: Calculate the limit as n-> infinity

How? Forgive my innocence.

4. ## Re: Calculate the limit as n-> infinity

Originally Posted by superduper1
How? Forgive my innocence.
$n^{\frac{2}{3}} * n^{\frac{2}{3}} = n^{\frac{4}{3}} = (n^{\frac{2}{3}})^2$ So Square root that to get $n^{\frac{2}{3}}$

5. ## Re: Calculate the limit as n-> infinity

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

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

What should I do next?

6. ## Re: Calculate the limit as n-> infinity

Basically, if you take a positive expression under a square root, you double the exponent. In general:

$a^b=\sqrt[n]{a^{bn}}$ where $0\le a^b$ if $n$ is even and $b$ is odd.

7. ## Re: Calculate the limit as n-> infinity

Originally Posted by superduper1
$\lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

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

What should I do next?
I would put the denominator also under the radical and use the property:

$\lim_{x\to c}\sqrt{f(x)}=\sqrt{\lim_{x\to c}f(x)}$

8. ## Re: Calculate the limit as n-> infinity

solutions says it is 1/2. I keep getting 1/infinity

9. ## Re: 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.

10. ## Re: 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.

11. ## Re: Calculate the limit as n-> infinity

Originally Posted by superduper1
$\lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$
How does the n^2/3 distribute?

$\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1}\le \frac{n^{\frac{2}{3}}n^{\frac{1}{2}}(\sqrt{2})}{2n ^2+n+1}$

Now use comparison.