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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 $$a_n = \frac{n^2}{\sqrt{n^3+9n}}$$


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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
wait for 4 minutes – dato datuashvili Oct 28 '12 at 8:31
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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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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