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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!

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    $\begingroup$ It's a good idea! $\endgroup$
    – Nosrati
    Aug 27, 2017 at 8:33
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    $\begingroup$ Did you encounter any problem? $\endgroup$
    – Carsten S
    Aug 27, 2017 at 9:25
  • $\begingroup$ Rewriting this as $\sqrt{4x-x^2} = \sqrt{4-(x^2-4x+4)} = \sqrt{4-(x-2)^2}$ might also help. $\endgroup$ Aug 27, 2017 at 13:29

4 Answers 4

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

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

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  • $\begingroup$ It would have been better if you assumed $|x-2|<1$ and then get $\delta =\min(1,2\epsilon) $. One should avoid complicated expressions for $\delta$ which smack of algebraic manipulation of inequalities. $\endgroup$
    – Paramanand Singh
    Aug 27, 2017 at 13:06
  • $\begingroup$ is this comlicated? $\endgroup$
    – Khosrotash
    Aug 27, 2017 at 13:20
  • $\begingroup$ Well I would say it is not complicated. I just added you could have made it simpler (and OP was also expecting something like $\delta=\min(.,.) $). $\endgroup$
    – Paramanand Singh
    Aug 27, 2017 at 13:23
  • $\begingroup$ You are right . I am thankful for your comment . $\endgroup$
    – Khosrotash
    Aug 27, 2017 at 13:24
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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.

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

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 $[0,4]$.

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  • $\begingroup$ Thanks! Could you explain the last step? $\endgroup$
    – Yuval Gat
    Aug 27, 2017 at 8:42
  • $\begingroup$ @Orangalo sorry, fixed. $\endgroup$
    – Kenny Lau
    Aug 27, 2017 at 9:13
  • $\begingroup$ This is not the way things work. Definition of limit is not an exercise in the algebraic manipulation of inequalities. For example your answer does not make sense if $\varepsilon=5$. $\endgroup$
    – Paramanand Singh
    Aug 27, 2017 at 13:04
  • $\begingroup$ @ParamanandSingh Indeed I assumed that $\varepsilon < 2$, because otherwise $\delta=2$ would suffice. You might want to point out why my approach is wrong apart from the unsubstantiated claim "definition of limit is not an exercise in the algebraic manipulation of inequalities" $\endgroup$
    – Kenny Lau
    Aug 27, 2017 at 13:06
  • $\begingroup$ The definition does not require one to solve the inequality $|f(x) - L|<\epsilon$ to find all values of $x$ for which the inequality holds. Unfortunately most cheap textbooks also follow this algebraic manipulation of inequalities. The method fails miserably for simple examples like $\lim_{x\to a} x^{2}=a^{2}$. $\endgroup$
    – Paramanand Singh
    Aug 27, 2017 at 13:13

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