# Prove the limit by using the epsilon delta proof

$$\lim_{x \to 0} x^2 = 0$$

My attempt,

Given $\epsilon>0,\exists \delta>0$ if

$$|x^2-0|<\epsilon \space \text{if} \space 0<|x-0|<\delta$$

$$|x|<\sqrt{\epsilon} \space \text{if} \space 0<|x|<\delta$$

This statement holds true if $\delta=\sqrt{\epsilon}$

Is my attempt correct? If it isn't, how to solve it? Thanks in advance.

Let $\epsilon > 0$. We wish to find a $\delta$ such that if $|x-0|<\delta$, then $|x^2-0|<\epsilon$.
Since $|x^2-0|=|x|^2$, we see that if $|x-0| = |x|<\sqrt\epsilon$, then $|x|^2=|x^2-0|<\epsilon$. Therefore, we set $\delta=\sqrt{\epsilon}$, and it follows from $|x-0|<\delta$ that $|x^2-0|<\epsilon$, which concludes the proof.