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

Question

[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:

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

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 $

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Is this proof correct?

Answer 1

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