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?
 A: Hint:
$$
\frac{1}{\sin x}-\frac{1}{\tan x} = \frac{1}{\sin x}-\frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x}\frac{x}{\sin x}.
$$
A: 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.
A: Answer:
$$\begin{align} 
\lim \limits_{x\to 0}\left(\dfrac{1}{\sin(x)}-\dfrac {1}{\tan (x)}\right)&=\lim \limits_{x\to 0}\left(\dfrac{1}{\sin(x)}-\dfrac {\cos(x)}{\sin(x)}\right)\\
&=-\lim \limits_{x\to 0}\left(\dfrac{\cos(x)-1}{\sin(x)}\right)\\
&=-\lim \limits_{x\to 0}\left(\dfrac{\cos(x)-\cos(0)}{x-0}\dfrac {x}{\sin(x)}\right)\\
&=-\lim \limits_{x\to 0}\left(\dfrac{\cos(x)-\cos(0)}{x-0}\right)\lim \limits_{x\to 0}\left(\dfrac {x}{\sin(x)}\right)\\
&=-\cos'(0)\cdot 1\\
&=-\sin(0)\\
&=0.
\end{align}$$
A: One way to do it, at least not mentioning L'Hospital would be
$$ \frac{1}{\sin x} - \frac{1}{\tan x} =
\frac{1-\cos x}{\sin x} = \frac{\frac12 x^2 + o(x^2)}{x + o(x^2)} $$
as $x\to 0$...
A: THE ANSWER
\begin{align}
\lim_{x\to0}\left(\frac{1}{\sin x}-\frac{1}{\tan x}\right)&=\lim_{x\to0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)\\
&=\lim_{x\to0}\left(\frac{1-\cos x}{\sin x}\right)\\
&=\lim_{x\to0}\left(\frac{1-\cos x}{\sin x}\cdot\frac{1+\cos x}{1+\cos x}\right)\\
&=\lim_{x\to0}\left(\frac{1-\cos^2 x}{\sin x(1+\cos x)}\right)\\
&=\lim_{x\to0}\left(\frac{\sin^2 x}{\sin x(1+\cos x)}\right)\\
&=\lim_{x\to0}\left(\frac{\sin x}{1+\cos x}\right)\\
&=\frac{\sin 0}{1+\cos 0}\\
&=\LARGE0
\end{align}
