# Calculate the limit as n-> infinity

• Nov 24th 2012, 07:44 PM
superduper1
Calculate 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 PM
MarkFL
Re: Calculate the limit as n-> infinity
$\displaystyle n^{\frac{2}{3}}=\sqrt{n^{\frac{4}{3}}}$
• Nov 24th 2012, 08:22 PM
superduper1
Re: Calculate the limit as n-> infinity
How? Forgive my innocence.
• Nov 24th 2012, 08:30 PM
jakncoke
Re: Calculate the limit as n-> infinity
Quote:

Originally Posted by superduper1
How? Forgive my innocence.

$\displaystyle n^{\frac{2}{3}} * n^{\frac{2}{3}} = n^{\frac{4}{3}} = (n^{\frac{2}{3}})^2$ So Square root that to get $\displaystyle n^{\frac{2}{3}}$
• Nov 24th 2012, 08:40 PM
superduper1
Re: 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 PM
MarkFL
Re: 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 PM
MarkFL
Re: Calculate the limit as n-> infinity
Quote:

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

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

What should I do next?

I would put the denominator also under the radical and use the property:

$\displaystyle \lim_{x\to c}\sqrt{f(x)}=\sqrt{\lim_{x\to c}f(x)}$
• Nov 24th 2012, 09:22 PM
superduper1
Re: Calculate the limit as n-> infinity
solutions says it is 1/2. I keep getting 1/infinity
• Nov 24th 2012, 09:50 PM
MarkFL
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.
• Nov 24th 2012, 10:37 PM
MarkFL
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.
• Nov 25th 2012, 04:30 AM
Plato
Re: Calculate the limit as n-> infinity
Quote:

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

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

Now use comparison.