Help me solve $\lim\limits_{x \to 0}\frac{2x\cos{2x}-\sin{2x}}{2x\cos{2x}-2x}$ First, I am pretty confident that the answer of this question must be like following:
$$ \lim_{x \to 0}\frac{2x\cos{2x} - \sin{2x}}{2x\cos{2x}-2x} = \lim_{x \to 0}\frac{\cos{2x} - \frac{\sin{2x}}{2x}}{\cos{2x}-1} = \lim_{x\to 0}\frac{\cos{2x}-1}{\cos{2x}-1} = 1 $$
but then someone told me that my solution is wrong since i perform partially limit in $\frac{\sin{2x}}{2x}=1$. So then i try so many other algebraic manipulation but got nothing. I need to solve this without using L-hospital, and that drive me crazy since i think it is the only method that i think will work on this. Please someone help me!
 A: One way to do it is by using that
$$\sin u=u+\frac{u^3}{6}+\mathcal{O}(u^5)$$
as $u\to 0$ and that
$$\cos u=1+\frac{u^2}{2}+\mathcal{O}(u^4)$$
as $u\to0$. Then the limit becomes
$$\lim_{x\to0}\frac{2x\cos 2x-\sin 2x}{2x\cos 2x-2x}=\lim_{u\to0}\frac{u\cos u-\sin u}{u\cos u-u}=\lim_{u\to0}\frac{u\left(1+\frac{u^2}{2}+\mathcal{O}(u^4)\right)-u-\frac{u^3}{6}+\mathcal{O}(u^5)}{u\left(1+\frac{u^2}{2}+\mathcal{O}(u^4)\right)-u}=\lim_{u\to0}\frac{\frac{1}{3}u^2+\mathcal{O}(u^5)}{\frac{1}{2}u^2+\mathcal{O}(u^5)}=\lim_{u\to0}\frac{\frac{1}{3}+\mathcal{O}(u^3)}{\frac{1}{2}+\mathcal{O}(u^3)}=\frac{2}{3}.$$
You could probably also do it using L'Hôpital's rule if you so wish, and perhaps there is some nice squeeze that I'm not seeing immediately.
A: We have,
\begin{align*}
\frac{2x\cos(2x)-\sin(2x)}{2x\cos(2x)-2x}&=\frac{t\cos(t)-\sin(t)}{t\cos(t)-t},\quad t=2x\not=0,\\
&=\frac{\frac{\cos(t)-1}{t^{2}}+\frac{t-\sin(t)}{t^{3}}}{\frac{\cos(t)-1}{t^{2}}},\\
&=\frac{-\frac{1}{2}+\frac{1}{6}}{-\frac{1}{2}},\quad t\to 0,\\
&=\frac{2}{3}.
\end{align*}
Here, we are using the fact
$$\lim_{t\to 0}\frac{\cos(t)-1}{t^{2}}=-\frac{1}{2},\quad \lim_{t\to 0}\frac{t-\sin(t)}{t^{3}}=\frac{1}{6}.$$
A: In fact,
\begin{align*}
&\frac{2x\cos(2x)-\sin(2x)}{2x\cos(2x)-2x}\\
&=\cos(2x)\frac{2x-\tan(2x)}{2x(\cos(2x)-1)}\\
&=\cos(2x)\frac{\frac{2x-\tan(2x)}{(2x)^3}}{\frac{\cos(2x)-1)}{(2x)^2}}
\end{align*}
Since
$$ \lim_{x\to0}\frac{x-\tan(x)}{x^3}=-\frac13, \lim_{x\to0}\frac{\cos(x)-1)}{x^2}=-\frac12$$
So
$$ \lim_{x\to0}\frac{2x\cos(2x)-\sin(2x)}{2x\cos(2x)-2x}=\frac23. $$
