The task is to find
$$
L=\lim _{n \rightarrow \infty} n^{2}-\frac{n}{\sin \left(\frac{1}{n}\right)}
$$
For sufficiently large $n$ we have $\sin \left(\frac{1}{n}\right) \approx \frac{1}{n}$ so that
$$
L \approx n^{2}-\frac{n}{\frac{1}{n}}=n^{2}-n^{2}=0
$$
so the limit $L$ is zero. Am I correct?
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$$
\sin \left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{6 n^{3}}+\frac{1}{120 n^{5}}+O\left(\frac{1}{n^{7}}\right)
$$
Now long division
$$
\begin{array}{l}
\frac{1}{\sin \left(\frac{1}{n}\right)}=n+\frac{1}{6 n}+\frac{7}{360 n^{3}}+O\left(\frac{1}{n^{5}}\right) \\
n^{2}-\frac{n}{\sin \left(\frac{1}{n}\right)}=-\frac{1}{6}-\frac{7}{360 n^{2}}+O\left(\frac{1}{n^{4}}\right)
\end{array}
$$
which shows the limit and how it is approached.
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