Question: I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:

$$\lim\_{ x\to \pi/4 } \frac{1 - \tan(x)}{x-\frac{\pi}{4}}$$

I know how to make it using L'Hopital and that the result is $-2$ ,but I'm getting nowhere when I try without it. Any advice?

Answer: 
We have the identities

$$\begin{align}\frac{1-\tan x}{x-\pi/4}&=-\frac{\tan(x-\pi/4)}{x-\pi/4}(1+\tan x)\\\\
&=-\frac{\sin(x-\pi/4)}{x-\pi/4}\frac{1+\tan x}{\cos(x-\pi/4)}
\end{align}$$

Can you finish from here?
You could use the following hint. Try to write $\tan(x)$ in terms of $\sin(x) $ and $\cos(x)$. A little bit of rearranging and then use the following,

$$\lim\_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$

A little bit of work yields the result.