Evaluation of trignometric limit I want to find the following limit without L'Hospital. 
$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$
Maybe I should try to get rid of the radical.

We will use the fact that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This will help us rationalize the numerator. With $a=1$and $b=(\tan x)^{1/3}$ we can write
\begin{align*}
\frac{1+(\tan x)^{1/3}}{1-2\cos^2x}& = \frac{1+(\tan x)^{1/3}}{1-2\cos^2x} \cdot \left[\color{blue}{\frac{1-(\tan x)^{1/3}+(\tan x)^{2/3}}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}}\right]\\
& = \frac{1+\tan x}{1-2\cos^2x}\cdot \left[\frac{1}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}\right]
\end{align*}
Now use the following identity:
$$\cos 2x=2 \cos^2 x -1 =\frac{1-\tan^2x}{1+\tan^2x},$$
to get
\begin{align*}
\frac{1+(\tan x)^{1/3}}{1-2\cos^2x}&= \frac{1+\tan^2 x}{\tan x-1}\cdot \left[\frac{1}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}\right]
\end{align*}
Now use the fact that $x \to 3\pi/4$ to get the limit as $-1/3$.
