Prove $\lim_{x\to0} \sqrt {4 − x} = 2$ using $ε$, $δ$-definition.

Question

Prove $\lim_{x\to0} \sqrt {4 − x} = 2$ using $\varepsilon$, $\delta$-definition.

Here's what I got:

Given $\varepsilon > 0$, choose a particular $\delta$.

Then whenever $0 < |x| < \delta$,

$|\sqrt {4 − x} - 2|$ has to equal $\varepsilon$ if we want to prove this limit.

I know that $4-x > 0$. How do I prove and present this question?

Thanks.

Answer 1

$y:= 4-x$, then $y \ge 0,$ for $x \le 4.$

Show : $\lim _{y \rightarrow 4} √y = 2$.

Let $\epsilon \gt 0$ be given.

Choose: $\delta =2\epsilon$, then:

$0 \le |√y -2| = |√y-2| \dfrac{√y+2}{√y+2} =$

$\dfrac{|y-4|}{√y +2} \le \dfrac{|y-4|}{2} \lt$

$\delta/2 = \epsilon.$

Answer 2

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