Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
@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
2  
@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 $\[-1,1\]$](math.stackexchange.com/questions/4764/…) –  Martin Sleziak Nov 7 '12 at 12:54

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.