Limit of trigonometric function with powers Find 
$$\lim_{x \to \frac{3\pi}{4}}\frac{1+(\tan x)^{1/3}}{1-2\cos^2x}$$
I don't think series expansion will help, then how do I solve it?

HINT:
$$1-2\cos^2x=-(2\cos^2x-1)=-\cos2x$$
and
$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$
and using $a^3-b^3=(a-b)(a^2+ab+b^2),$
$$1+(\tan x)^{1/3}=\frac{1+\tan x}{1+(\tan x)^{1/3}+(\tan x)^{2/3}}$$
For $x\to\dfrac{3\pi}4,\tan x\to\tan\dfrac{3\pi}4=\tan\left(\pi-\dfrac{\pi}4\right)=-\tan\dfrac{\pi}4=-1$
$\implies\tan x\ne-1\iff\tan x+1\ne0,$ so can be cancelled out safely
