# I know the value of the limit is &#x2212;<!-- - --> <mstyle displaystyle="true" scriptlevel="0"

I know the value of the limit is $-\frac{1}{6}$. I tried to use the fact
$\mathrm{ln}\left(ab\right)=\mathrm{ln}\left(a\right)+\mathrm{ln}\left(b\right)$
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kawiarkahh
We use
$\underset{y\to 0}{lim}\frac{\mathrm{ln}\left(y+1\right)}{y}=1$
Letting $y=n\mathrm{sin}\frac{1}{n}-1$ so
${n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)={n}^{2}\left(n\mathrm{sin}\frac{1}{n}-1\right)\frac{\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)}{n\mathrm{sin}\frac{1}{n}-1}$
so the limit is replaced by
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\left(n\mathrm{sin}\frac{1}{n}-1\right)$
and as remarked this is an application of
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x-x}{{x}^{3}}$