# Prove $x^2$ is continuous on the interval $[0,1]$

I realize this may be a simple question, but I am having trouble proving proving this, mostly with selecting a suitable delta.

My attempt so far:

$x^2$ is continuous on $[0,1]$ if $\forall\epsilon>0,\exists\delta>0$ such that $$0<\vert x-a\vert<\delta\,x\in [0,1]\Rightarrow\vert x^2-a^2\vert<\epsilon$$ for all $a\in [0,1]$.

Since $a$ and $x$ are in $[0,1]$, the maximum value of $a+x$ is 2. Take $\delta=\min\{1,\frac{\epsilon}{2}\}$. Then $$0<\vert x-a\vert<\delta\Rightarrow\vert x^2-a^2\vert=\vert x-a\vert\vert x+a\vert<2\vert x-a\vert=2\delta\leq\epsilon$$

I think my logic makes sense but I am unsure about my step involving $\delta$. Is this a correct proof? If not, how would I improve it?

• Looks good to me! Now I have a question for you: why take $\delta$ to be the minimum of $1$ and $\epsilon/2$? Why not just let it be $\epsilon/2$? – layman Oct 23 '15 at 4:57
• In the end you need to get a strict inequality according to the definition So you might want to make $\delta = \epsilon /4$ – happymath Oct 23 '15 at 4:58
• This proof is perfect and is exactly how a continuity proof is supposed to go! My only thought is as $x^2$ is continuous on unbound R, can you do this without using the max value of a + x? – fleablood Oct 23 '15 at 5:04
• Oooh, the "less than 2*...." should be a less than or equal. Very minor but as it need to be a strict inequality.... – fleablood Oct 23 '15 at 5:11

The only improvement I can see here (as mentioned by happymath in the comments), is that the implication requires strict inequality. So, choosing something like $\delta=\frac{\epsilon}{3}$ yields a better result. This gives the implication:
$$2\delta=\frac{2\epsilon}{3}<\epsilon$$
This is the required condition. Of course, as you probably know, $\delta=\frac{\epsilon}{3}$ is a somewhat arbitrary choice. You can choose anything less than $\frac{\epsilon}{2}$ (i.e. $\delta=\frac{\epsilon}{2.0000000001}$, or $\delta=\frac{\epsilon}{123456789}$ ). Or, more generally, as proposed in the comments, consider $\delta<\min(1,\frac{\epsilon}{2})$.
• Or he could simply say, "let $\delta <\min(1 , \epsilon/2)$" – fleablood Oct 23 '15 at 5:07
• It doesn't hurt that $\delta$ is chosen as the minimum of $1$ and $\epsilon/2$, but how does it help in this case? Why not just take $\delta = \epsilon/2$? – layman Oct 23 '15 at 5:13
• I took $\delta=\min\{1,\epsilon/2\}$ because when I was initially working out the problem, I set $\delta=1$ and noted that $1<x-a<1\Rightarrow -(x+a)<x^2-a^2<x+a\leq2$, and then I modified my definition of $\delta$ from that. But I didnt have a firm idea of what I was doing – Matt G Oct 23 '15 at 5:28