Prove $f(x) = x^2$ is continuous at $x=4$ I want to show that $f(x) = x^2$ is continuous at $x=4$
and here's how the proof goes:
$\forall\epsilon>0$, $\exists\delta>0$ s.t $\forall x$, $|x-4|<\delta$ $\implies |f(x)-16|<\epsilon$
So working backwards we get:
$$|f(x)-16|<\epsilon ⇔ |x^2 - 16| < \epsilon$$
                     $$⇔ |x+4||x-4| < \epsilon$$
Ideally we're trying to get it into this form: $|x-4|<\delta$
Since we can't divide $\epsilon$ by $|x+4|$ we need to control that term by assuming a priori that $\delta \le 1$
Here's where I have few questions:


*

*How do you control a term

*What are we ultimately trying to get by controlling the term by assuming a priori that $\delta \le 1$ (how's it going to help us proceed?)


If anyone can answer the above question as if I'm completely new to the idea of continuity and $\epsilon-\delta$ that would be really appreciated.
I think if I'm able to understand how to control a term and the purpose behind it(how's it going to help us in the end) than I think I might be able to finish the proof on my own.
Note: I don't want the complete proof for this question, but rather if someone can answer those questions above.

To control the factor $|x+4|$ we use the triangle inequality and the fact that $|x-4|<\delta$ so we  have
$$|x|-4\le|x-4|<\delta\Rightarrow |x|<4+\delta$$
and by assuming that $\delta\le1$ we find
$$|x+4|\le |x|+4<8+\delta\le9$$
hence
$$|x-4||x+4|<9|x-4|<\epsilon$$
and then we choose $\delta=\min\left(\frac\epsilon9,1\right)$ and the game is played.
