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[Proof-verification] Determining whether the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$; $x_0\in \mathbb{R}$ is continous $\color{red}{\text{ in }x_0}$ or not with the $\varepsilon$-$\delta$-definition of limit/criterion:


$$f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$$

Proof:
Let $\varepsilon>0$ and ${\mid x-x_0\mid}<\delta \iff {\mid x-3\mid}<\delta$

\begin{align} {\mid f(x)-f(x_0)\mid}&= {\mid\sqrt{x^2-x-6}-\sqrt{3^2-3-6}\mid}\\ & ={\mid\sqrt{x^2-x-6}\mid}\\ & ={\mid\sqrt{x+2} \cdot \sqrt{x-3}\mid}\\ & <\sqrt{x+2}\cdot \sqrt{x+2}\\ & = x+2\\ & < x-3+5 \\ & <\delta+5 =: \varepsilon \\ & \iff \delta = \varepsilon -5 \end{align}

$\implies$ the function is continous in $x_0=3 \qquad \qquad \qquad \qquad \qquad_\blacksquare $


Is this proof correct?

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  • $\begingroup$ So, if $\varepsilon = 1$, what's the $\delta > 0$? $\endgroup$ – Ennar Nov 6 '18 at 15:54
  • $\begingroup$ @Ennar It wouldn't be positive $\endgroup$ – Doesbaddel Nov 6 '18 at 18:07
  • $\begingroup$ And that's not really good now, is it? $\endgroup$ – Ennar Nov 6 '18 at 18:23
  • $\begingroup$ No, because $\delta$ needs to be positve, right? $\endgroup$ – Doesbaddel Nov 6 '18 at 18:28
  • $\begingroup$ Exactly.$\hphantom{}$ $\endgroup$ – Ennar Nov 6 '18 at 18:35
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First of all, the function $f$ is not defined on $\mathbb{R}$ but for $$ x\in(-\infty,-2]\cup[3,+\infty). $$

So one can only talk about its continuity of $f$ at $x=3$ from the right.

For $x>3$, you are right to get $$ |f(x)-f(3)|=\sqrt{x-3}\sqrt{x+2}. $$ Note that you don't need the absolute value for the square root terms.

But then you made a mistake: the $\delta$ you get must be positive.

The term $\sqrt{x-3}$ should not be dropped and it would give you the desired $\delta$.

Consider instead for $0<x-3<1$ the inequality $$ \sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}\le\varepsilon. $$

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  • $\begingroup$ Can you explain more in detail how you came to this conclusion and the estimation upwards? $\Big($Consider instead for $0<x-3<1$ the inequality $\sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}\le\varepsilon\Big)$. $\endgroup$ – Doesbaddel Nov 6 '18 at 18:25
  • $\begingroup$ Why is $\sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}$? With your hint I would come to the following conclusion: $\cdots \leq 6 \cdot \sqrt{x-3} < 6 \sqrt{ \delta }=: \varepsilon \implies \delta = \dfrac{\varepsilon^2}{36}$ Is that correct? $\endgroup$ – Doesbaddel Nov 6 '18 at 19:13
  • $\begingroup$ @Doesbaddel: If $0<x-3<1$, then $3<x<4$ and thus $\sqrt{x+2}<\sqrt{6}<6$. I just did a rough estimate. $\endgroup$ – user587192 Nov 6 '18 at 21:38
  • $\begingroup$ @Doesbaddel: you would need $\delta=\min\{1,\frac{\varepsilon^2}{36}\}$. $\endgroup$ – user587192 Nov 6 '18 at 21:39
  • $\begingroup$ Why does $x-3$ needs to be between $0$ and $1$? Where does this step come from? Why do you need $\delta=\min\{1,\frac{\varepsilon^2}{36}\}$ and not $\delta= \frac{\varepsilon^2}{36}$? (Otherwise, your estimation makes sense now and I understood it.) $\endgroup$ – Doesbaddel Nov 6 '18 at 21:53

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