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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
    
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, How does accept rate work?. –  Did Oct 28 '12 at 8:26
    
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

3 Answers 3

up vote 1 down vote accepted

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!

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2  
I think you meant $+\infty$... –  Johnny Westerling Oct 28 '12 at 8:26
    
I apologise for that! –  Host-website-on-iPage Oct 28 '12 at 8:33

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 $-\infty$.

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