Limit of trigonometric function with powers

Question

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?

Answer 1

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

Answer 2

Alternatively, the limit can be transformed into a rational expression through the substitution $tanx=t^3$ From here you can deduce (through right triangle trigonometry) the relation $cos^2x=\frac{1}{t^6+1}$ The expression becomes after simplification $\frac{(1+t)(t^6+1)}{t^6-1}$ which can be reduced through factoring as Lab indicated in his answer. When $x$ goes to 3pi/4, you can figure out what $t$ does. Now there is less to worry about trig stuff