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.
 A: 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:
$$|f(x)-16| < 9|x-4| < 9\left(\frac{\epsilon}{9}\right) = \epsilon$$
A: 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.
A: We want to show that $|(x-4)(x+4)|$ can be made "small" by choosing $x$ suitably close to $4$. So the size of $|x-4|$ is under our control. The term $x+4$ could spoil things. However, if we say that in any case we will choose $|x-4|\lt 1$, then we know that $|x+4|$ will be less than $9$.  
A: Notice that $|x-4|<\delta$ is equivalent to $4-\delta < x < 4+\delta$. Thus, if we add $4$ in every term of this inequality, we get $8-\delta<x+4<8+\delta$, so that $|x+4|<8+\delta$. Since we eventually want $|x+4||x-4|<\varepsilon$, we want $(8+\delta)(\delta)<\varepsilon$. If $\delta<1$, then $(8+\delta)(\delta)<9\delta$, and so $\delta<\min\{\varepsilon/9,1\}$ will do the trick.
Now we can write the whole proof:
Let us show that $f(x)=x^2$ is continuous at $x=4$. Let $\varepsilon>0$ be arbitrary. Then, there is $\delta=\min\{\varepsilon/9,1\}>0$ such that for all $|x+4|<\delta$ we have
$$|f(x)-f(4)|=|x^2-16|=|x+4||x-4|< (8+\delta)(\delta)\leq 9\delta=9\min\{\varepsilon/9,1\}\leq 9\cdot \varepsilon/9=\varepsilon,$$
where we have used the fact that if $|x-4|<\delta$, then $|x+4|<8+\delta$, and the fact that $\delta<1$ and $\delta<\varepsilon/9$. Hence $|f(x)-f(4)|<\varepsilon$ for all $x$ with $|x-4|<\delta$, and the proof is finished.
