find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} 
   \ -\sqrt{2x+2}}$$ without l'hospital rule. 
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} 
   \ -\sqrt{2x+2}} = -6$.
I can show my attempt but it's pointless as the result i achieved $+\infty$, which is very wrong. is there an strategy trying to calculate a limit without l'hospital rule.

Starting from J.C.Santos result 
$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$
I'd multiply num and den by $\left(2x+\sqrt{x^2+3}\right)$ to get
$$\lim_{x\to1}\frac{\left(4x^2-x^2-3\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$
that is
$$\lim_{x\to1}\frac{3(x+1)(x-1)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$
and then
$$-\lim_{x\to1}\frac{3(x+1)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{\left(2x+\sqrt{x^2+3}\right)}=-\frac{24}{4}=-6$$
