# limit computation $\lim_{n\rightarrow \infty}\frac{\ln (1+n^{3})-\ln(n^{6})}{\sin ^{3}(n)}$.

Help me please to compute this limit: $\displaystyle \lim_{n\rightarrow \infty}\frac{\ln (1+n^{3})-\ln(n^{6})}{\sin ^{3}(n)}$.

Thanks a lot!

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@Manzano That is not true. It does not vanish for any $n\in \Bbb N$. $\sin x=0\iff x=k\pi \;,k\in\Bbb Z$. –  Pedro Tamaroff Nov 5 '12 at 12:41
@PeterTamaroff: Maple says it doesn't exist. Are we supposed to search two sequences for violating the limit? –  Babak S. Nov 5 '12 at 13:03
@BabakSorouh It doesn't exist because the logs go to $-\infty$ but the sine oscilates infinitely often. –  Pedro Tamaroff Nov 5 '12 at 13:04
So true, Peter. –  Babak S. Nov 5 '12 at 13:07
About the character of the sequence $(\sin n)$: Sine function dense in $$-1,1$$ –  Martin Sleziak Nov 7 '12 at 12:54

We have $$\frac{\ln (1+n^3) - \ln (n^6)}{\sin^3 n} = \frac{\ln n}{\sin^3 n} \left(\frac{\ln(1+n^3)}{\ln n} - \frac{\ln(n^6)}{\ln n}\right).$$ The expression in the right brackets tend to $-3$. Thus it suffices to consider the limit $$\lim_{n\to \infty} \frac{\ln n}{\sin^3 n}.$$ Since $\ln n\to \infty$ and $0<|\sin n|\leqslant 1$, we get that in absolute value the limit is infinity. However, as $\sin^3 n$ changes signs infinitely often, the limit does not exist.