Prove $\lim_{x\to2} \sqrt{4x-x^2} = 2$ by definition with $\epsilon, \delta$

Question

I need to prove using $\epsilon, \delta$:

$$\lim_{x\to2} \sqrt{4x-x^2} = 2$$

Meaning I need to prove that for every $\epsilon > 0$, there is a $\delta >0$ such as that $0<|x-2|<δ→|f(x)-2|<ϵ$

I tried going like this: $|\sqrt{4x-x^2} -2|<|\sqrt{4x}-2|<|2\sqrt x-2|=|2(\sqrt x-1)|=2|\sqrt x-1|<2|\sqrt x-1|$...

But I don't know how to choose my $\delta$ from here.

Answer 1

Hint: $$ \left|\sqrt{4x-x^2}-2\right|= \left|\frac{4x-x^2-4}{\sqrt{4x-x^2}+2}\right|= \frac{(x-2)^2}{\sqrt{4x-x^2}+2}\le\frac{1}{2}(x-2)^2 $$

Answer 2

Egreg's method is the cleanest, but here is another approach for variety's sake:

Choose $\epsilon>0$ and note that $x \mapsto \sqrt{x}$ is increasing. Then if $\eta = \min(4-(2-\epsilon)^2, (2+\epsilon)^2-4)$ and $|x-4| < \eta$, then $|\sqrt{x} -2| < \epsilon$.

Now we try to find a $\delta$ so that if $|x-2| < \delta$, then $|4x-x^2 -4| < \eta$. Since $|4x-x^2 -4| = (x-2)^2$, we see that if $\delta = \sqrt{\eta}$ and $|x-2| < \delta$, then $|4x-x^2 -4| < \eta$ and hence $|\sqrt{4x -x^2} - 2| < \epsilon$.

So, we can take $\delta = \sqrt{\min(4-(2-\epsilon)^2, (2+\epsilon)^2-4)}$.