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: 
Recall that
$$
f'(a) = \lim\_{x\to a}\frac{f(x) - f(a)}{x-a}.
$$
Apply it to the case where $f(x) = \tan x$ and $a=\pi/4$. Then $f(a) = 1$ and $f'(a) = \sec^2 a = \sec^2(\pi/4) = 2$. Therefore
$$
2 = \lim\_{x\to\pi/4}\frac{\tan x - \tan(\pi/4)}{x-\pi/4}.
$$

So $-2$ is the answer to the question as you've posed it.
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.