Calculating $\lim_{x \to 0}\frac{x \tan2x-2x\tan x}{(1-\cos2x)^2}$ I'm a bit confused regarding this question. I've been trying to solve it and have gotten to the same answer ($3/8$) thrice now. I have no idea where I'm going wrong and would really appreciate some help figuring it out. Here's my solution:
$$\lim_{x \to 0}\frac{x\tan2x-2x\tan x}{(1-\cos2x)^2}$$
Applying L'Hospital's rule,
$$\lim_{x \to 0}\frac{\tan2x+x(\sec^22x)(2)-2\tan x-2x\sec^2x}{2(1-\cos2x)(\sin2x)(2)}$$
Dividing numerator and denominator by $2x$,
$$\begin{align*}
\lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\
&=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\
&=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)}
\end{align*}$$
Applying L'Hôpital's rule (again),
$$\lim_{x \to 0}\frac{(2)(\sec2x)(\sec2x)(\tan2x)(2)-(2)(\sec x)(\sec x\tan x)}{4(\sin2x)(2)}$$
$$=\lim_{x \to 0}\frac{(2)(\sec^22x)(\tan2x)-(\sec^2x)(\tan x)}{4(\sin2x)}$$
Dividing numerator and denominator by $2x$,
$$\begin{align*}
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{2x})}{4(\frac{\sin2x}{2x})}\\
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{x})(\frac{1}{2})}{4(\frac{\sin2x}{2x})}\\
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(1)-(\sec^2x)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{(2)(\sec^20)(1)-(\sec^20)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{(2)(1)(1)-(1)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{2-\frac{1}{2}}{4}\\
&=\frac{\frac{3}{2}}{4}\\
&=\frac{3}{8}.
\end{align*}$$
 A: Your problem arises here:
\begin{align*}
\lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\
&=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\
&=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)}
\end{align*}
particularly where you evaluate $\lim_\limits{x\to 0} \frac {\tan 2x}{2x} = 1$ and $\lim_\limits{x\to 0} -\frac {2\tan x}{2x} = -1$ and cancel them.
$\tan 2x - 2\tan x\approx 2x^3$ and that is not trivial.
A: It's a small conceptual mistake after applying L'Hospital's rule for the 1st time then "dividing Nr and Dr by 2x and then applying the limits individually without expressing limit as a sum."
Do note that after dividing the Nr and Dr with 2x and then applying the limits individually you get a completely new function not same as the one before i.e. line 2 and line 3 are two different function.
You can still solve this without much hard work.. just express everything in terms of tanx..
A: Hint:
Another way:
Observe that
$$x\tan2x-2x\tan x=x(\tan2x-2x)-2x(\tan x-x)$$
Use
Are all limits solvable without L'Hôpital Rule or Series Expansion to find $$\lim_{h\to0}\dfrac{\tan h-h}{h^3}=\dfrac13$$
Can you take it from here?
A: I am now going to find the limit without L’Hospital Rule.
$$\begin{array}{l}
 \displaystyle \quad \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{\frac{x \sin 2 x}{\cos 2 x}-\frac{2 x \sin x}{\cos x}}{(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin 2 x \cos x-2 x \sin x \cos 2 x}{\cos 2 x \cos x(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x\left(\sin 2 x \cos x-\sin  x \cos 2 x\right)-x \sin x \cos 2 x}{\tan 2 x \cos x(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin x(1-\cos 2 x)}{\cos 2 x \cos x(1-\cos 2 x)^{2}}\\
=\displaystyle \left(\lim _{x \rightarrow 0} \frac{x \sin x}{1-\cos 2 x}\right)\left(\lim _{x \rightarrow 0} \frac{1}{\cos 2 x \cos x}\right)\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin x}{2 \sin ^{2} x}\\
=\displaystyle \frac{1}{2} \lim _{x \rightarrow 0} \frac{x}{\sin x}\\
=\displaystyle \frac{1}{2}
\end{array}$$
A: Hint:
Use Taylor-Young's formula at (the final) order $4$ and do some trigonometry:

*

*$\tan x= x+\dfrac{x^3}3+o(x^3)$;

*$(1-\cos2x)^2=(2\sin^2 x)^2=4\sin^4x=4\bigl(x+o(x)\bigr)^4$.

