Show continuity of a function using epsilon delta definition

Question

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!!

Answer 1

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

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