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?

• Also I suspect you'll need to use Taylor series, can you use them? – user88595 Apr 22 '14 at 12:26

Hint: $$\frac{1}{\sin x}-\frac{1}{\tan x} = \frac{1}{\sin x}-\frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x}\frac{x}{\sin x}.$$
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 $x\to 0$...
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}
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}