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: 
**Hint:**

Let $t=x-\frac{\pi}{4}$, then
$$\frac{1-\tan x}{x-\frac{\pi}{4}}=\frac{1-\tan\left(t+\frac{\pi}{4}\right)}{t}=\frac{1}{t}\cdot\left(1-\frac{\tan t+\tan\frac{\pi}{4}}{1-\tan \frac{\pi}{4}\tan t}\right)=\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)=\frac{-2\tan t}{t(1-\tan t)}$$
Now, take the limit as $t\to 0$:
\begin{align}
\lim\_{x\to \frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}&=\lim\_{t\to 0}\frac{-2\tan t}{t(1-\tan t)}\\
&=\lim\_{t\to 0}\frac{-2\tan t\cos t}{t(1-\tan t)\cos t}\\
&=-2\lim\_{t\to 0}\frac{\sin t}{t(\cos t-\sin t)}\\
&=-2\left(\lim\_{t\to 0}\frac{\sin t}{t}\right)\left(\lim\_{t\to 0}\frac{1}{\cos t-\sin t}\right)\\
&=-2(1)(1)\\
&=\color{blue}{-2}
\end{align}
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.