[Math] Evaluating $\lim_{x\to 0}\left(\frac{1}{\sin x} – \frac{1}{\tan x}\right)$

Question

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?

Answer 1

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}