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I have a homework problem that I arrived.

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With Mathematica, the limit is 0. So by using $\epsilon= 10^{-6}$ (it is -6, not -0, sorry for the cutoff).

$\sin(n^2)/\sqrt{n} <\epsilon =10^{-6}$

So I tried putting that ino Mathematica and no luck so I have a feeling I am approaching this problem the wrong way

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Hint: $\frac{sin(n^{2})}{\sqrt{n}}\leq\frac{1}{\sqrt{n}}$ and the latter is monotonic so it suffices to find an $N$ s.t $\frac{1}{\sqrt{N}}<\epsilon$

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How do you know if the sequence $1/\sqrt{n}< \epsilon$ in the first place (the same epsilon)? –  sidht Sep 22 '12 at 23:59
    
Find such $N$. Hint: find $N$ s.t $\sqrt{N}>\frac{1}{\epsilon}$ (why ?) –  Belgi Sep 23 '12 at 0:01
    
You should get $10^{12} = N$ –  sidht Sep 23 '12 at 0:03
    
see my edit to the comment, I had a typo –  Belgi Sep 23 '12 at 0:06
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We have $|\sin \theta|\le 1$ for all $\theta$, so if we can make $\frac{1}{\sqrt{n}}\lt \epsilon$, then for sure $\left|\frac{\sin(n^2)}{\sqrt{n}}\right|\lt \epsilon$. –  André Nicolas Sep 23 '12 at 1:39
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