How do I do this limit proof

Prove using the precise definition of the limit:

lim x^2 = 0
x-> 0

Let ε>0. Then there exists a δ>0 such that |x|<δ -> |x^2|<ε. So |x^2| < ε suggests that x
GUESS: Take δ = sqrt(ε).

PROOF:

Let ε>0 and δ = sqrt(ε).

Suppose that 0<|x|<δ. Then
|f(x) - 0| = |x^2 - 0| = |x^2| ...... then I don't know what to do next! I need to show that it's lesser than δ, correct? But how do I even do that?

Is it something like |x^2| < δ^2 = (sqrt(ε))^2 = ε?

Thank you.

Saying |x^2| < δ^2 would be the next step, but if you're getting confused, you can write it out slightly differently. I notice that you frequently start with |f(x) - L|, and attempt to prove inequalities in one line. This time, let's work from the assumption that x lies in the delta-neighbourhood:

0 < |x - 0| < δ
===> |x| < sqrt(ε)
===> |x|^2 < sqrt(ε)^2 ... (squaring is an increasing function when dealing with positive numbers, so no sign change is necessary)
===> |x^2| < ε ... (remember that absolute values distribute over multiplication)
===> |x^2 - 0| < ε
===> |f(x) - 0| < ε

You could have done this in one line if you wanted to, but this way might help get it more straight in your mind.