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.

Multiplying the top and the Bottom by $ \left(2x+\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right) $, gives the following :
\begin{aligned} \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=\lim_{x\to 1}{\frac{-3\left(x+1\right)\left(\sqrt{2x+2}+\sqrt{x+3}\right)}{\sqrt{x^{2}+3}+2x}}\\ &=\frac{-3\left(1+1\right)\times\left(\sqrt{2\times 1+2}+\sqrt{1+3}\right)}{\sqrt{1^{2}+3}+2\times 1}\\ \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=-6 \end{aligned}
