<s>[INST] 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!
 [/INST] You were right to multiply "top" and "bottom" by the conjugate of the numerator. I suspect you simply made a few algebra mistakes that got you stuck with the limit you first posted:
So we start from the beginning:
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$ 
and multiply top and bottom by the conjugate, $\;\sqrt{x^2 + 3} + 2$:
$$\lim_{x \to1} \, \frac{\sqrt{x^2 + 3} -2 }{x^2 - 1} \cdot \frac{\sqrt{x ^ 2 + 3} +2}{\sqrt{x^2+3}+2}$$
You were correct to do that. You just miss-calculated and didn't actually need to expand the denominator, that's all. 
In the numerator, we have a difference of squares (which is the reason we multiplied top and bottom by the conjugate), and expanding the factors gives us: $$(\sqrt{x^2+3}-2)(\sqrt{x^2 + 3} + 2) = (\sqrt{x^2+3})^2 - (2)^2 = (x^2 + 3) - 4 = x^2 - 1$$ And now there's no reason to waste time trying to simplify the denominator*, since we can now cancel the factor $(x^2 - 1)$ from top and bottom:
$$\lim_{x\to1}\frac{\color{blue}{\bf (x^2- 1)}}{\color{blue}{\bf (x^2 - 1)}(\sqrt{x^2 +3}+2)}\; = \;\lim_{x \to 1} \dfrac{1}{\sqrt{x^2 + 3}+2}\; = \;\frac{1}{\sqrt{1+3} + 2} \;=\; \dfrac 14$$
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