The question gave me this limit to proof the result by using epsilon-delta:
$\lim_{x\to 0} \sqrt{4-x} = 2$
It seems to be using conjugate multiplication, but I got lost and I didn't know how to proceed in order to find $\delta$.
If its possible, show me your equation developtment please.
Ps.: My problem here is just to find $\delta$, I know what to do next.
All help is appreciated and adored!
$|\sqrt{4-x}-2|=|\frac{(\sqrt{4-x}-2)(\sqrt{4-x}+2)}{\sqrt{4-x}+2}|=|\frac{x}{\sqrt{4-x}+2}|$
Note that if $|x|\leq 1$ then $\sqrt{4-x}+2\geq \sqrt{3}+2$ and in that case we get $|\sqrt{4-x}-2|\leq\frac{|x|}{\sqrt{3}+2}$. We want that expression to be at most $\epsilon$, and it happens when $|x|\leq\epsilon(\sqrt{3}+2)$. Since we also had to assume $|x|\leq 1$ we take $\delta=\min\{1,\ \epsilon(\sqrt{3}+2)\}$.