What is the value of $\lim_{x\to0}\frac{2\cot 2x-\cot x}{\sin 2x}$ ?
$1)\text{zero}\qquad\qquad2)\frac12\qquad\qquad3)-\frac12\qquad\qquad4)-1$
Here is my work:$$\lim_{x\to0}\frac{2\cot 2x-\cot x}{\sin 2x}=\lim_{x\to0}\left(\frac{2\cos2x}{\sin^22x} -\frac{\cos x}{\sin x\sin 2x}\right)$$Second fraction can be simplified to $\frac{\cos x}{2\sin^2 x\cos x}=\frac{1}{2\sin^2x}$. But when $x\to0$ I get $\infty-\infty$
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$\begingroup$A helpful double-angle identity for this problem is $\cot 2x=\frac{\cot x - \tan x}2$. With that established, $$\frac{2\cot 2x-\cot x}{\sin 2x}=\frac{-\tan x}{2\sin x\cos x}=-\frac{\sin x/\cos x}{2\sin x\cos x}=-\frac1{2\cos^2x}$$ The RHS is defined at $x=0$, leading to a limit of $-\frac12$.
I hadn't seen that elegant identity before, so I'm grateful that Google was closer than my copy of Schaum's Mathematical Handbook. ^_^ Just to demonstrate it: $$\cot 2x=\frac{\cos2x}{\sin2x}=\frac{\cos^2x-\sin^2x}{2\sin x\cos x}=\frac{\cos x}{2\sin x}-\frac{\sin x}{2\cos x}=\frac{\cot x-\tan x}2$$
$\endgroup$ $\begingroup$The way you've proceeded also will get you to the correct answer. Continuing from the point where you left :$$L=\lim_{x\to 0} \frac{2\cos(2x)}{\sin^2(2x)}-\frac{1}{2\sin^2(x)}$$Use the identity $\cos (2x)=1-2\sin^2(x)$ gives$$\lim_{x\to 0} \frac{2}{\sin^2(2x)}-\frac{4\sin^2(x)}{\sin^2(2x)}-\frac{1}{2\sin^2(x)}=\lim_{x\to 0} \frac{1}{2\sin^2(x)\cos^2 (x)}-\underbrace{\frac{1}{\cos^2(x)}}_{=1}-\frac{1}{2\sin^2(x)}=$$Now we evaluate the left out part :$$L=\lim_{x\to 0} \frac{1}{2\sin ^2(x)\cos^2(x)}-\frac{1}{2\sin^2(x)}-1=\lim_{x\to 0} \frac{\overbrace{1-\cos^2(x)}^{=\sin^2(x)}}{2\sin^2(x)\cos^2(x)}-1$$$$=\lim_{x\to 0} \frac{1}{2\cos^2(x)}-1 = \frac{1}{2}-1=-\frac{1}{2}$$
$\endgroup$ $\begingroup$Using abbreviations, I continue from where you left off.$${2(2c^2-1)\over4s^2c^2}-{c\over2s^2c} ={2c^2-1-c^2\over2s^2c^2}$$$$={-s^2\over2s^2c^2}={-1\over2c^2}\,\to-{1\over2}$$
$\endgroup$ 1 $\begingroup$If you don't find elegant double angle formulas, you can try and work out the numerator, because it's in $\infty-\infty$ form.
You have $\cot2x=1/\tan2x$ and therefore$$ 2\cot2x-\cot x=2\dfrac{1-\tan^2x}{2\tan x}-\dfrac{1}{\tan x}=-\tan x $$so your fraction becomes$$ -\frac{\sin x}{\cos x}\frac{1}{2\sin x\cos x}=-\dfrac{1}{2\cos^2x} $$and the conclusion follows.
$\endgroup$ $\begingroup$Note that\begin{align}\lim_{x\to0}\frac{\cot(x)}x-\frac1{x^2}&=\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}\\&=\lim_{x\to0}\frac{\left(x-\frac12x^3+\frac1{4!}x^5-\cdots\right)-\left(x-\frac16x^3+\frac1{5!}x^5-\cdots\right)}{x^3-\frac13x^5+\cdots}\\&=\lim_{x\to0}\frac{-\frac13x^3+\cdots}{x^3+\cdots}\\&=-\frac13.\end{align}Therefore$$\lim_{x\to0}\frac{2\cot(2x)}x-\frac1{x^2}=4\lim_{x\to0}\frac{\cot(2x)}{2x}-\frac1{(2x)^2}=-\frac43.$$It follows that$$\lim_{x\to0}\frac{2\cot(2x)-\cot(x)}x=-\frac43-\left(-\frac13\right)=-1,$$and so$$\lim_{x\to0}\frac{2\cot(2x)-\cot(x)}{\sin(2x)}=\lim_{x\to0}\frac{\frac{2\cot(2x)-\cot(x)}x}{\frac{\sin(2x)}x}=\frac{-1}2=-\frac12.$$
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