# Determine that the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$ is continous with the $\varepsilon$-$\delta$-definition of limit/criterion

[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?

• So, if $\varepsilon = 1$, what's the $\delta > 0$? – Ennar Nov 6 '18 at 15:54
• @Ennar It wouldn't be positive – Doesbaddel Nov 6 '18 at 18:07
• And that's not really good now, is it? – Ennar Nov 6 '18 at 18:23
• No, because $\delta$ needs to be positve, right? – Doesbaddel Nov 6 '18 at 18:28
• Exactly.$\hphantom{}$ – Ennar Nov 6 '18 at 18:35

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 $$0 the inequality $$\sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}\le\varepsilon.$$

• 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 $\sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}\le\varepsilon\Big)$. – Doesbaddel Nov 6 '18 at 18:25
• 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? – Doesbaddel Nov 6 '18 at 19:13
• @Doesbaddel: If $0<x-3<1$, then $3<x<4$ and thus $\sqrt{x+2}<\sqrt{6}<6$. I just did a rough estimate. – user587192 Nov 6 '18 at 21:38
• @Doesbaddel: you would need $\delta=\min\{1,\frac{\varepsilon^2}{36}\}$. – user587192 Nov 6 '18 at 21:39
• 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.) – Doesbaddel Nov 6 '18 at 21:53