# How do you prove the statement lim as x approaches 0 for x^2 = 0 using the epsilon and delta definition?

Oct 9, 2015

See the explanation, below.

#### Explanation:

You need to show that if we are given a positive number that we'll call $\epsilon$, then there is a number $\delta$ (also positive) that makes the following true:

if $x$ is chosen so the $0 < \left\mid x - 0 \right\mid < \delta$, then $\left\mid {x}^{2} - 0 \right\mid < \epsilon$.

One way to do this is to notice that if $\left\mid x \right\mid < \sqrt{\epsilon}$, the $\left\mid x \right\mid < \epsilon$

Choose $\delta = \sqrt{\epsilon}$

Another way is to observe that for $\left\mid x \right\mid < 1$, we have $\left\mid {x}^{2} \right\mid < \left\mid x \right\mid$. Choose $\delta = \min \left\{1 , \epsilon\right\}$

Whichever way is chosen, we then write up the proof:

Proof

Given $\epsilon > 0$, choose $\delta = \text{whatever}$

Now if $0 < \left\mid x - 0 \right\mid < \delta$, then

[insert a proof that $\left\mid x \right\mid < 0$ implies $\left\mid {x}^{2} \right\mid < \epsilon$]