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!
 A: 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
$$\lim_{x \rightarrow 1}    \ \frac{\sqrt{x^2+3}-2}{x^2 - 1} \ \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2} $$
$$ = \ \lim_{x \rightarrow 1} \frac{(x^2+3) \ - \ 4 }{(x^2 - 1) \ \cdot \ (\sqrt{x^2+3}+2) } = \ \lim_{x \rightarrow 1} \frac{x^2 \ - \ 1 }{(x^2 - 1) \ \cdot \ (\sqrt{x^2+3}+2) } $$
$$= \ \lim_{x \rightarrow 1} \frac{ 1 }{ \sqrt{x^2+3}+2 } \ = \ \frac{1}{4} \ . $$
A: 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:
$$\lim_{x\to1}\frac{\color{blue}{\bf (x^2- 1)}}{\color{blue}{\bf (x^2 - 1)}(\sqrt{x^2 +3}+2)}\; = \;\lim_{x \to 1} \dfrac{1}{\sqrt{x^2 + 3}+2}\; = \;\frac{1}{\sqrt{1+3} + 2} \;=\; \dfrac 14$$
A: 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
$$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{x^2-1}=\lim_{t\to 0}\frac{\sqrt{t^2+2t+4}-2}{t^2+2t}=\lim_{t\to 0}2\frac{\sqrt{\frac{t^2+2t}{4}+1}-1}{t^2+2t}=\lim_{t\to 0}2\frac{t/4}{2t}=\frac{1}{4}$$
A: 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...) $\lim_{x\to 1}\frac{\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{2x}=\frac{\frac{1}{2}(1^2+3)^{-\frac{1}{2}}(2(1))}{2(1)}=\frac{\frac{1}{2}(4)^{-\frac{1}{2}}(2)}{2}=\frac{1}{2}(4)^{-\frac{1}{2}}=(\frac{1}{2})(\frac{1}{4})^{-\frac{1}{2}}=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$
