<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] This limit problem seems to cooperate rather nicely with the "conjugate factor" method.  Multiply the numerator and denominator by the "conjugate" of the numerator to obtain
$$\lim_{x \rightarrow 1}    \ \frac{\sqrt{x^2+3}-2}{x^2 - 1} \ \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2} $$
$$ = \ \lim_{x \rightarrow 1} \frac{(x^2+3) \ - \ 4 }{(x^2 - 1) \ \cdot \ (\sqrt{x^2+3}+2) } = \ \lim_{x \rightarrow 1} \frac{x^2 \ - \ 1 }{(x^2 - 1) \ \cdot \ (\sqrt{x^2+3}+2) } $$
$$= \ \lim_{x \rightarrow 1} \frac{ 1 }{ \sqrt{x^2+3}+2 } \ = \ \frac{1}{4} \ . $$
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