Evaluate the following limit
$$\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}$$
From wolfram alpha, answer is $-\frac{1}{3}$. From here, we obtain the answer by using either Taylor series or L'Hopital rule. I try to apply L'Hopital rule to this question. I obtain
$$\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}=\lim_{n \rightarrow \infty}{\frac{n^2\sin^2(\frac{1}{n})-1}{\sin^2(\frac{1}{n})}} =\frac{-1}{0}$$
What's wrong with my working?