Proving continuity of $x^{2}-2$ with $\epsilon$ and $\delta$ Here is my attempt:
$f$ is continuous in $p$ if
$\forall \epsilon > 0\; \exists \delta > 0$ such that $|x-p|<\delta \Rightarrow |f(x) - f(p)|<\epsilon$
Then for $f(x)=x^{2}-2$ in $x=p$
$|f(x) - f(p)|<\epsilon$
$|x^2-2 - p^2+2|$
= $|x^2-p^2|$
= $|x^2-p^2+2xp-2xp|$
=$|(x-p)^2+2xp|$
I can't figure out how to proceed from this point. I tried explore the triangle inequality, but apparently it would not lead me to something helpful.
 A: The idea is to bound $|x-p|$ by something. Let $\epsilon > 0$ and suppose $|x-p| < 1$. From this, we have: $$|x|-|p| \leq |x-p| < 1 \implies |x| < 1+|p|.$$
We want to find $\delta > 0$ such that $|f(x)-f(p)| < \epsilon$ if $|x-p| < \delta$. We have: $$\begin{align} |f(x)-f(p)| &= |x²-2-(p^2-2)| \\ &= |x^2 - p^2| \\ &= |x+p||x-p| \\ &\leq (|x|+|p|)|x-p| \quad \Delta \text{ inequality} \\ &< (1+|p|+|p|) |x-p|\quad \text{extra hypothesis} \\ &=(1+2|p|)\delta\end{align}$$
If I want to make this $< \epsilon$, then $\delta = \epsilon/(1+2|p|)$ would be nice. However, I need the hypothesis $\delta < 1$ so that I have $|x| < 1+|p|$ to make all these steps valid. So $\delta = \min\left\{1, \frac{\epsilon}{1+2|p|}\right\} > 0$ will work. You might find my answer here also helpful.
A: Hint: We have $x^2-p^2=(x-2)(x+2)$.  Assume that $\delta$ is fairly small (say $\delta\leq 1$).  From this can you get a bound on the $x+2$ term? In the end, take a new $\delta'$ that takes the above assumption into consideration.  
