<s>[INST] 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.
 [/INST] As the OP suggested openness to asymptotic analysis, we proceed accordingly.  
First, we note the expansion of the numerator is given by
$$\begin{align}
1+\tan^{1/3} x&=1+\left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\
&=\frac23 (x-3\pi/4)+O(x-3\pi/4)^2 \tag 1
\end{align}$$ 
whereas the expansion of the denominator is given by
and 
$$\begin{align}
1-2\cos^2x&=-\cos 2x\\\\
&=-2(x-3\pi/4)+O(x-3\pi/4)^3 \tag 2
\end{align}$$
Thus, putting $(1)$ and $(2)$ together yields 
$$\begin{align}
\frac{1+\tan^{1/3} x}{1-2\cos^2x}&=\frac{\frac23 (x-3\pi/4)+O(x-3\pi/4)^2}{-2(x-3\pi/4)+O(x-3\pi/4)^3}\\\\
&=-\frac13+O(x-3\pi/4)\\\\
&\to -\frac13\,\,\text{as}\,\,x\to 3\pi/4
\end{align}$$
Therefore, we have that the limit of interest is 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 3\pi/4}\frac{1+\tan^{1/3} x}{1-2\cos^2x}=-\frac13}$$

NOTES:
To arrive at $(1)$, we first note that for $f(x)=\tan x$, $f'(x)=\sec^2 x$ and $f''(x)=2\sec^2 x\tan x$.  
Therefore, $f(3\pi/4)=-1$ and $f'(3\pi/4)=\frac12$ so that
$$\tan x=-1+2(x-3\pi/4)+O(x-3\pi/4)^2$$
Then, using the expansion for $(-1+x)^{1/3}=-1+\frac13x+O(x^2)$, we see that 
$$\begin{align}
\tan^{1/3} x&= \left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\
&=-1+\frac23(x-3\pi/4)+O(x-3\pi/4)^2
\end{align}$$

To arrive at $(2)$, we first note the trigonometric identity $\cos 2x=2\cos^2-1$.  
Then, for $g(x)=1-2\cos^2 x=-\cos 2x$, we have $g'(x)=2\sin 2x$, $g''(x)=4\cos 2x$, and $g'''(x)=-8 \sin 2x$.  
Therefore, $g(3\pi/4)=0$ and $g'(3\pi/4)=-2$, and $g''(3\pi/4)=0$ so that
$$-\cos 2x=-2(x-3\pi/4)+O(x-3\pi/4)^3$$
</s>