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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?

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

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

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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 $x\to 0$...

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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.

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

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