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.

If you assume $\delta < 1$, then you know that:
$$\begin{align*}
|x-4|<\delta \\
\Rightarrow |x-4|<1 \\
\Rightarrow -1 < x-4 < 1 \\
\Rightarrow 3 < x < 5
\end{align*} $$
But then, we can determine what this means about $|x+4|$:
$$\begin{align*}
&\Rightarrow 7 < x+4 < 9 \\
&\Rightarrow |x+4| < 9
\end{align*} $$
So this means that if we assume $\delta<1$, we have: 
$$|f(x)−16| = |x+4||x-4| < 9|x-4|$$
And if you let $\delta = \min\left(1,\frac{\epsilon}{9}\right)$, then:
$$|f(x)-16| < 9|x-4| < 9\left(\frac{\epsilon}{9}\right) = \epsilon$$
