Prove $f(x) = x^2$ is continuous at $x=4$

Question

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:

1. How do you control a term
2. 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.

Answer 1

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:

$$|f(x)-16| < 9|x-4| < 9\left(\frac{\epsilon}{9}\right) = \epsilon$$

Answer 2

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.

Answer 3

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 $9$.

Answer 4

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.