Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$ 
Thanks!

Let $t=x-1$ so $x=t+1$ and since $(1+y)^\frac{1}{2}\sim_0 1+\frac{y}{2}$ and $y^2=_0o(y)$ then we find
$$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{x^2-1}=\lim_{t\to 0}\frac{\sqrt{t^2+2t+4}-2}{t^2+2t}=\lim_{t\to 0}2\frac{\sqrt{\frac{t^2+2t}{4}+1}-1}{t^2+2t}=\lim_{t\to 0}2\frac{t/4}{2t}=\frac{1}{4}$$
