$\varepsilon - \delta$ proof that $f(x) = x^2 - 2$ is continuous - question concerning the initial choice of $\delta$

Question

I just realized I did non really internalized $\varepsilon - \delta$ proofs. Here there is an attempt, with some general questions I have.

Proposition: $f(x) := x^2 - 2$ is continuous.

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Skratch Work: We assume that $|x - c| < \delta$. Thus, we have

\begin{align} | x^2 - 2 - c^2 + 2 | & = | x^2 - c^2 | \\ & = | x-c| \cdot |x+c|. \end{align}

Assume that $\delta = \frac{c}{2}$. Then we have

$$ |x - c | < \frac{c}{2} \Longleftrightarrow \frac{c}{2} < x < \frac{3c}{2} .$$

By adding $c$ we obtain

$$\frac{3c}{2} < x +c < \frac{5c}{2} .$$

Hence, $x + c < \frac{5c}{2}$, which implies that

$$ | x - c | \cdot | x+ c| < \frac{5c}{2} | x - c| < \varepsilon,$$

if $\delta = \frac{2\varepsilon}{5c}$.

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Proof: Let $c$ be arbitrary, and assume that $| x - c | < \delta$. Let $\delta := \min \{ \frac{c}{2} , \frac{2\varepsilon}{5c} \}$. Hence, \begin{align} | x^2 - 2 - c^2 + 2 | & = | x^2 - c^2 | \\ & = | x-c| \cdot |x+c| < \frac{5c}{2} | x - c| < \frac{5c}{2} \delta = \varepsilon. \end{align}

Being $c$ arbitrary, this establishes the continuity of $f(x)$. $\square$

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Questions:

1. Is this proof correct?

2. I chose the initial value of $\delta$ arbitrarily, but in such a way that depends on the initial choice of $c$ (in doing so I did not follow a proof of this proposition I found, that starts by taking $\delta = 1$). I did this by following a well-known proof of the continuity of $f(x) = \frac{1}{x}$.
   Is this sound?

Answer 1

Your proof didn't cover the case $c < 0$. I would do it as follows:Let $c \in \mathbb{R}, \epsilon > 0$ be arbitrary, choose $\delta = \min \{1,\frac{\epsilon}{1+2|c|}\}$. Thus if $|x-c| < \delta $, then $$|x-c||x+c| \leq |x-c|(|x-c| + 2|c|)< |x-c|(1+2|c|)< \dfrac{\epsilon}{1+2|c|}\cdot (1+2|c|) = \epsilon .$$