Show that $\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}=-\frac{1}{3}$ 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?

Use substitution $y:=\frac{1}{x}.$ Then $x\to+\infty \implies y\to0.$ So, $$\lim_{n\to+\infty}\Biggl(n^2-\frac{1}{\sin^2\bigl(\frac{1}{n}\bigr)}\Biggr)=\lim_{y\to0}\Biggl(\frac{1}{y^2}-\frac{1}{\sin^2(y)}\Biggr) = \lim_{y\to0}\Biggl(\frac{\sin^2(y)-y^2}{y^2\sin^2(y)}\Biggr).$$ Applying L'Hopital rule 4 times yields the following limit: $$\lim_{y\to0}\Biggl(\frac{\sin^2(y)-y^2}{y^2\sin^2(y)}\Biggr) = \lim_{y\to0}\Biggl(\frac{-4\cos(2y)}{12\cos(2y)-16y\sin(2y)-4y^2\cos(2y)}\Biggr).$$ Since $$\begin{align} -4\cos(2y)\rightarrow -4, && (y\to0) \\ 12\cos(2y)\rightarrow12, && (y\to0) \\ -16y\sin(2y)\rightarrow0, && (y\to0) \\ -4y^2\cos(2y)\rightarrow 0, && (y\to0) \end{align}$$ the last limit is equal to $$\lim_{y\to0}\Biggl(\frac{-4\cos(2y)}{12\cos(2y)-16y\sin(2y)-4y^2\cos(2y)}\Biggr) = -\frac{4}{12} = -\frac{1}{3}.$$
