Precise definition of limits and $\lim_{x\to-1}\frac1{\sqrt{x^2+3}}=\frac12$

Question

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!

Answer 1

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.