Prove. $\lim_ {x \to 0 } \frac{x^2}{|x|} = 0$. From first principles.

Question

I'm a student taking a real analysis course at university and I have been asked the following question. I'm still learning so if anyone has any tips they would be much appreciated!

I start by fixing $\epsilon > 0$ and find $\delta > 0$ such that.

$$0 < |x - 0| < \delta \implies \bigg| \frac{x^2}{|x|} - 0 \bigg| < \epsilon$$

$$0 < |x| < \delta \implies \bigg| \frac{x^2}{|x|} \bigg| < \epsilon$$

$$ \impliedby \frac{|x^2|}{|x|} < \epsilon$$

$$ \impliedby \frac{|x||x|}{|x|} < \epsilon$$

$$ \impliedby |x| < \epsilon$$

$$0 < |x| < \delta \implies |x| < \epsilon$$

Therefore take $\delta = \epsilon$.

Thanks for your time!

Answer 1

Fix $\epsilon > 0.$ Choose $\delta = \epsilon.$

Observe that if $0 < |x - 0| < \delta,$ then we have $\bigg| \frac{x^2}{|x|} - 0 \bigg|= \bigg| \frac{x^2}{|x|} \bigg |=|x| < \delta=\epsilon.$

Therefore, $\delta = \epsilon$ works, and we are done.

I just rephrased your work.

Answer 2

The point is that $\epsilon-\delta$-proofs are found by working backwards, but the end-product must be a forward proof.

Your argument contains the essence of a correct proof, but really the end product should be something like this:

Fix $\epsilon > 0$. If $0 < |x-0| < \epsilon$, then

$$\left|\frac{x^2}{|x|}-0\right| = |x| < \epsilon$$