# Show continuity of a function using epsilon delta definition

I have the following question:

Show that the function $f(x) = x^2$ is continuous at every $a \in R$ by >using the definition of continuity (i.e., show that for every $\epsilon > 0$ there is a $\delta > 0$ such >that $|f(x) - f(a)| <\epsilon$ whenever $|x-a|< \delta$)

That means that for every $\epsilon > 0$ there is a $\delta > 0$ such that $|x^2-a^2|< \epsilon$ whenever $|x-a|< \delta$. I honestly have no idea how to even start that, so help would be very much appreciated!!

• Try to use the identity $x^2 - y^2 = (x - y)(x + y)$. – TheGeekGreek Oct 8 '16 at 14:36
• So I get $|(x-a)(x+a)|<\epsilon$ whenever $|x-a|<\delta$. So if I combine them I get $|x-a|<\delta<|(x-a)(x+a)|<\epsilon$. But how can I show that there exists such a $\delta$ for every $\epsilon$? – Maria Oct 8 '16 at 15:04

assume $a\geq0$. let $\epsilon$ be a real $>0$.

we have for every real $x$

$x^2-a^2=(x-a)(x+a)$

we now assume that $x$ is such that

$|x-a|<\color{red}{1}$ for example since we are near the point $x=a$,

which means

$a-1<x<a+1$ and $-2a-1\leq 2a-1<x+a \leq 2a+1$ thus $|x+a|\leq 2a+1$

finally

$|x^2-a^2| \leq (2a+1)|x-a|$

and if

$|x-a|<\frac{\epsilon}{2a+1}$

then

$|x^2-a^2|<\epsilon$

you could take

$\delta=min(\color{red}{1}$ $,\frac{\epsilon}{2a+1})$

• From where does this come from: $-2a-1\le2a-1<x+a\le2a+1$? – Maria Oct 8 '16 at 15:16
• i assumed $|x-a|<1$ . we are near a. you could impose if you want $|x-a|<2$ – hamam_Abdallah Oct 8 '16 at 15:18
• I'm afraid I don't understand, so you have just assumed $|x-1|<1$ because x is close to a so their distance will be a small number, right? I understand that you got from that $a-1<x<a+1$. But I still don't understand the next step, why do you assume $|x-a|<2$? – Maria Oct 8 '16 at 15:21
• as you are close to a, you impose what you want $|x-a|<1$ or $<2$ or |<3\$... – hamam_Abdallah Oct 8 '16 at 15:24
• look at the red 1. – hamam_Abdallah Oct 8 '16 at 15:29