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.

let us make a change of variable $x = 3\pi/4 + h.$ then we have 
$$\begin{align}\tan(x) &= \tan(3\pi/4 + h) = \frac{-1 + \tan h}{1 + \tan h} \\
&= (\sin h - \cos h)(\sin h + \cos h \cdots)^{-1} \\
&=(-1 + h \cdots)(1+h +\cdots)^{-1} \\
&=(-1 + h \cdots)(1-h +\cdots) \\
&=-1+2h+\cdots\\
\tan^{1/3} x &=-1 + \frac23 h+\cdots\\
1+\tan^{1/3} x &= \frac23 h+\cdots \tag 1\\
\sqrt 2\cos x &= \sqrt 2\cos(3\pi/4+h)= -\cos h - \sin h\\
&=-1 - h + h^2/2+\cdots\\
2\cos^2 x &=  1+2h+\cdots\\
1-2\cos^2 x &= -2h+\cdots \tag 2
\end{align}$$
from $(1)$ and $(2),$  $$\lim_{x \to 3\pi/4} \frac{1+\tan^{1/3} x}{1-2\cos^2 x }=\frac{2h/3+\cdots}{-2h + \cdots} = -\frac 13. $$
