<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] Use L'Hospital's Rule. Since plugging in $x=1$, gives you indeterminate form, take the derivative of the numerator and the derivative of the denominator, and try the limit again.
$\lim_{x\to 1}\frac{(x^2+3)^{\frac{1}{2}}-2}{x^2-1}\implies$ (Via L
Hospital's Rule...) $\lim_{x\to 1}\frac{\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{2x}=\frac{\frac{1}{2}(1^2+3)^{-\frac{1}{2}}(2(1))}{2(1)}=\frac{\frac{1}{2}(4)^{-\frac{1}{2}}(2)}{2}=\frac{1}{2}(4)^{-\frac{1}{2}}=(\frac{1}{2})(\frac{1}{4})^{-\frac{1}{2}}=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$
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