<s>[INST] Precise definition of limits and $\lim_{x\to-1}\frac1{\sqrt{x^2+3}}=\frac12$ 
Prove using the epsilon-delta definition,
  $$
\lim_{x\to-1}\frac1{\sqrt{x^2+3}}=\frac12
$$

what I came up so far was:
$0<|x+1|<δ$ implies $\left|\frac1{\sqrt{x^2+3}} - \frac12\right|<ϵ$
\begin{align}
&\frac1{\sqrt{x^2+3}} - \frac12 \\
&=\frac{2\sqrt{x^2+3}-(x^2+3)}{2(x^2+3)} \\
&= \frac{2\sqrt{(x+1)^2-2(x+1)+4} - ((x+1)^2-2(x+1)+4)}{2[(x+1)^2-2(x+1)+4]} 
\end{align}
I'm unable to proceed because there are no real roots for $y^2-2y+4$
Can some please help me with this question? Thanks!
 [/INST] Perhaps it is not the best idea to make the denominator square free
\begin{align}
\frac{1}{\sqrt{x^2+3}}-\frac12 
&= \frac{2-\sqrt{x^2+3}}{2\sqrt{x^2+3}}=\frac{4-(x^2+3)}{2\sqrt{x^2+3}(2+\sqrt{x^2+3})}
\\
&=\frac{(1-x)(1+x)}{2\sqrt{x^2+3}(2+\sqrt{x^2+3})}
\end{align}
using binomial formulas twice.
</s>