Need help with epsilon delta proof that $\lim_{x\to 2}\sqrt{4x-x^2}=2$

Question

I need to prove the following statement using the epsilon-delta definition of a limit: $$\lim_{x\to 2}\sqrt{4x-x^2}=2$$

I tried to calculate $|f(x)-2|$ and tried to simplify it to $|x-2|g(x)$ so I could limit delta (probably ≤ 1) and then calculate a bound for $g(x)$, and then let $\varepsilon=\min\{1,\frac{\varepsilon}{bound}\}$

Thanks in advance!

Answer 1

Let $f(x) = \sqrt{4x-x^2}$.

We want to prove,using the $\epsilon$-$\delta$ method, that $\lim_{x\to 2}f(x)=2$.

Fix $\epsilon > 0$, and let $\delta=\min(2,\sqrt{2\epsilon})$.

Suppose $|x-2| < \delta$. Our goal is show $|f(x)-2| < \epsilon$.

Note that

$\qquad|x-2|<\delta \implies |x-2| < 2 \implies 0 < x < 4 \implies f(x) > 0$.

Also note that

$\qquad|(f(x)-2)(f(x)+2)| = |f(x)^2-4| = |4x-x^2-4|=|-(x-2)^2|=(x-2)^2$. \begin{align*} \text{Then}\;\;&|x-2| < \delta\\[4pt] \implies\;&|x-2| < \sqrt{2\epsilon}\\[4pt] \implies\;&|x-2|^2 < 2\epsilon\\[4pt] \implies\;&(x-2)^2 < 2\epsilon\\[4pt] \implies\;&|(f(x)-2)(f(x)+2)| < 2\epsilon\\[4pt] \implies\;&|f(x)-2||f(x)+2| < 2\epsilon\\[4pt] \implies\;&|f(x)-2|2 < 2\epsilon\\[0pt] &\qquad\text{[since $f(x) > 0 \implies |f(x) + 2| = f(x) + 2 >2$]}\\[4pt] \implies\;&|f(x)-2| < \epsilon\\[4pt] \end{align*} as was to be shown.

Answer 2

Hint: $$\sqrt{4x-x^2}-2=(\sqrt{4x-x^2}-2)\times\dfrac{\sqrt{4x-x^2}+2}{\sqrt{4x-x^2}+2}=\dfrac{-(x-2)^2}{\sqrt{4x-x^2}+2}$$

Answer 3

$$\begin{array}{rlcccl} & -\varepsilon &<& \sqrt{4x-x^2}-2 &<& \varepsilon \\ \iff& 2-\varepsilon &<& \sqrt{4x-x^2} &<& 2+\varepsilon \\ \iff& 2-\varepsilon &<& \sqrt{4x-x^2} &\le& 2 \\ \iff& (2-\varepsilon)^2 &<& 4x-x^2 &\le& 4 \\ \iff& \varepsilon^2-4\varepsilon &<& -4+4x-x^2 &\le& 0 \\ \iff& \varepsilon^2-4\varepsilon &<& -(x-2)^2 &\le& 0 \\ \iff& 0 &\le& (x-2)^2 &<& 4\varepsilon - \varepsilon^2 \\ \iff& 0 &\le& |x-2| &<& \sqrt{4\varepsilon - \varepsilon^2} \\ \end{array}$$

It is assumed that $4\varepsilon-\varepsilon^2 \ge 0$, i.e. $\varepsilon \le 2$.

If $\varepsilon > 2$, then one only need to set $\delta=2$, for the domain of definition of $f$ is just $[0,4]$.

Answer 4

another point of wiew$$\forall \epsilon>0 , \delta >0 :|x-2|<\delta \implies |\sqrt{4x-x^2}-2|<\epsilon\\ |\frac{4x-x^2-4}{\sqrt{4x-x^2}+2}|<\epsilon\\ |\frac{-(x-2)^2}{\sqrt{4-(x-2)^2}+2}|\leq |\frac{-(x-2)^2}{0+2}|<\epsilon\\\to |\frac{(x-2)^2}{2}|<\epsilon\\|(x-2)^2|<2\epsilon\\|\sqrt{4x-x^2}-2|\leq |x-2|<\sqrt{2\epsilon} $$