Evaluate the given limit: $\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$ Evaluate the given limit
$$\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
My Attempt :
$$=\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
$$=\lim_{x\to 0} \dfrac {x(\tan (2x)-2\tan (x))}{1-2\cos (2x)+\cos^2 (2x)}$$

With equivalents, there are less calculations:
First rewrite the expression as $\;\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}$.
Now,


*

*it's standard that $\;1-\cos u\sim_0\dfrac{u^2}2$, so $\;(1-\cos 2x)^2\sim_0(2x)^2 $.

*Taylor's formula at order $3$ for the tangent is $\;\tan u=u+\dfrac{u^3}3+o(u^3)$, so
$$\tan 2x-2\tan x=2x+\frac{8x^3}3 -\Bigl(2x+\frac{2x^3}3\Bigr)+o(x^3)=2x^3+o(x^3)\sim_0 2x^3, $$
and eventually
$$\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}\sim_0\frac{2x^4}{4x^4}=\frac12.$$

