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.

The expression tends to $$\frac{2\cdot1-\sqrt{1^2+3}}{\sqrt{1+3}+\sqrt{2\cdot1+2}}=0.$$
You may not use L'Hospital here.
(Or is there a typo in the question ?)

Update:
$$\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}=\frac{\sqrt{x+3}+\sqrt{2x+2}}{2x+\sqrt{x^2+3}}\cdot \frac{4x^2-x^2-3}{x+3-2x-2}\to\frac{4}{4}(-6).$$
