0
$\begingroup$

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!

$\endgroup$

closed as unclear what you're asking by Carsten S, José Carlos Santos, Leucippus, Siong Thye Goh, user223391 Aug 29 '17 at 15:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ It's a good idea! $\endgroup$ – Nosrati Aug 27 '17 at 8:33
  • 1
    $\begingroup$ Did you encounter any problem? $\endgroup$ – Carsten S Aug 27 '17 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. $\endgroup$ – Martin Sleziak Aug 27 '17 at 13:29
1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '17 at 13:06
  • $\begingroup$ is this comlicated? $\endgroup$ – Khosrotash Aug 27 '17 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 '17 at 13:23
  • $\begingroup$ You are right . I am thankful for your comment . $\endgroup$ – Khosrotash Aug 27 '17 at 13:24
0
$\begingroup$

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

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 $[0,4]$.

$\endgroup$
  • $\begingroup$ Thanks! Could you explain the last step? $\endgroup$ – Yuval Gat Aug 27 '17 at 8:42
  • $\begingroup$ @Orangalo sorry, fixed. $\endgroup$ – Kenny Lau Aug 27 '17 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 '17 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 '17 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 '17 at 13:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.