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.

I will assume that you meant$$\lim_{x\to1}\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}.$$Multiplying the numerator and the denominator by $\sqrt{x+3}+\sqrt{2x+2}$, this becomes$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$So, define $f(x)$ as $2x-\sqrt{x^2+3}$ and then the limit that you're after is just$$-f'(1)\left(\sqrt{1+3}+\sqrt{2+2}\right)=-4f'(1)$$(which is $-6$ indeed).
