Evaluating $\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)$ How to solve this limit
$$
\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)
$$
without using L'Hospital's rule?

By the formulas
$$
\cos 2x=\cos x ^2-\sin x ^2=1-2\sin x ^2, \sin 2x=2 \sin x \cos x,
$$
we have
$$
1-\cos x=2\sin^2 {\frac{x}{2}}, \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}.
$$
Then,
$$\begin{align}
\lim_{n\rightarrow \infty} (\frac{1}{\sin x}-\frac{1}{\tan x})
&=\lim_{n\rightarrow \infty} (\frac{1}{\sin x}-\frac{\cos x}{\sin x})\\
&=\lim_{n\rightarrow \infty} \frac{1-\cos x}{\sin x}\\
&=\lim_{n\rightarrow \infty} \frac{2\sin^2 {\frac{x}{2}}}{2 \sin \frac{x}{2} \cos\frac{x}{2}}\\
&=\lim_{n\rightarrow \infty} \frac{\sin {\frac{x}{2}}}{\cos\frac{x}{2}}\\
&=\frac{0}{1}\\
&=0.
\end{align}
$$
P.S. identities among triangle functions are very helpful when one wants to simplify formulas involving them.
