Given this limit: $\lim_{x->-1} \frac{\sqrt{x+5}-2}{x+1}=\frac{1}{4}$ Find a $\delta > 0 $ such that for $ |x-(-1)| < \delta $ exists:
$|\frac{\sqrt{x+5}-2}{x+1}-L|<\frac{1}{1000}$
I started with:
$|\frac{\sqrt{x+5}-2}{x+1}-\frac{1}{4} | < \epsilon$
which can be simplified like this:
$$ |\frac{(\sqrt{x+5}-2)(\sqrt{x+5}+2)}{(x+1)(\sqrt{x+5}+2)}-\frac{1}{4} | <\epsilon $$
$$ |\frac{x+1}{(x+1)(\sqrt{x+5}+2)}| < \epsilon $$
$$ |\frac{1}{(\sqrt{x+5}+2)}| < \epsilon$$
And got stuck again...I thought I should used the $x+1$ because I already know that $|x+1|<\delta$ but it disappeared with the calculation.. any suggestions?
\begin{align} \left|\dfrac{\sqrt{x+5}-2}{x+1}-\dfrac{1}{4}\right|&=\left|\dfrac{2-\sqrt{x+5}}{4(2+\sqrt{x+5})}\right|\\ &=\left|\dfrac{2-\sqrt{x+5}}{4(2+\sqrt{x+5})}\cdot\dfrac{2+\sqrt{x+5}}{2+\sqrt{x+5}}\right|\\ &=\left|\dfrac{-x-1}{4(2+\sqrt{x+5})^2}\right|\\ &=\dfrac{|x+1|}{4(2+\sqrt{x+5})^2} \end{align}
Note that $\dfrac{|x+1|}{4(2+\sqrt{x+5})^2}\leq\dfrac{|x+1|}{16}$ and $$\left|\dfrac{\sqrt{x+5}-2}{x+1}-\dfrac{1}{4}\right|=\dfrac{|x+1|}{4(2+\sqrt{x+5})^2}\leq\dfrac{|x+1|}{16}<\varepsilon\Rightarrow|x+1|<16\varepsilon$$
So we choose $\delta=16\varepsilon$ and we are done.