Finding the limit, multiplication by the conjugate I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$
I tried and tried... friends of mine tried as well and we don't know how to get out of:
$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$
(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)
How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.

You had the right idea: the issue is in your simplification of the numerator:
$$\begin{align}
(2 - \sqrt{3 + x})(2 + \sqrt{3 + x}) & = 2^2 - \left(\sqrt{(3 + x)}\right)^2 \\ \\
& = 4 - (3 + x) \\ \\
& = 4 - 3 - x \\ \\ 
& = 1 - x = -(x - 1)
\end{align}$$
That gives you $$\begin{align} \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(2 + \sqrt{3 + x}} &  = \lim_{x\to 1} \frac{-1}{2 + \sqrt{3 + x}}
& = -\frac 14
\end{align}$$
