# Can someone help me with this limit of sequence?

This is a question from my homework that I do not know how to approach it. Please help!

Find the limit of the sequence as n approaches infinity $$a_n = \frac{n^2}{\sqrt{n^3+9n}}$$

Thanks!

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Is there a reason for the 0% accept rate? –  Dennis Gulko Oct 28 '12 at 8:21
(1.) What did you try? (2.) What about this accept rate? –  Did Oct 28 '12 at 8:21
I have no idea about the 0% accept rate –  Jaden Q Oct 28 '12 at 8:23
I tried to click the check mark, but it keeps say "you can accept an answer in 4 minutes", what should i do then? –  Jaden Q Oct 28 '12 at 8:29
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let us make some operation, but please after find you answer accept it,first let us take out n out of brackets,we would have $n^2/\sqrt{n*({9+n^2})}$,after you get here,you can cancel out both side by $\sqrt{n}$,we get $n*\sqrt(n) /(\sqrt{9+n^2})$,take n in radical ,you get

$\sqrt{n^3}/(\sqrt{9+n^2})$,remember that $\sqrt{a/b}= \sqrt{a}/(\sqrt{b})$ and finally you get that answer is +infinity,because in numerator degree is bigger than in denumenator

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See that $$a_n=\frac{\sqrt n}{\sqrt{1+9/n^2}}$$ Clearly the limit is infintity!
I think you meant $+\infty$... –  Johnny Westerling Oct 28 '12 at 8:26
Informally and intuitively, limit to infinity can always be thought of as limit to some very large number. If n were very large $n^3$ would be much larger than $9n$, hence denominator would be approx $n^{1.5}$, hence whole fraction that is $a_n$ would be $n^{0.5}$ which would be infinitely large if n were infinitely large. Also sqaure root could be both +ve and -ve, so $a_n$ could be approaching both $+\infty$ and $-\infty$.