Evaluating $\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)$

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}

Hint: $$ \frac{1}{\sin x}-\frac{1}{\tan x} = \frac{1}{\sin x}-\frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x}\frac{x}{\sin x}. $$

Answer: $$\begin{align} \lim \limits_{x\to 0}\left(\dfrac{1}{\sin(x)}-\dfrac {1}{\tan (x)}\right)&=\lim \limits_{x\to 0}\left(\dfrac{1}{\sin(x)}-\dfrac {\cos(x)}{\sin(x)}\right)\\ &=-\lim \limits_{x\to 0}\left(\dfrac{\cos(x)-1}{\sin(x)}\right)\\ &=-\lim \limits_{x\to 0}\left(\dfrac{\cos(x)-\cos(0)}{x-0}\dfrac {x}{\sin(x)}\right)\\ &=-\lim \limits_{x\to 0}\left(\dfrac{\cos(x)-\cos(0)}{x-0}\right)\lim \limits_{x\to 0}\left(\dfrac {x}{\sin(x)}\right)\\ &=-\cos'(0)\cdot 1\\ &=-\sin(0)\\ &=0. \end{align}$$