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!
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!
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.