Use epsilon-delta definition to show $f(x)=x^2$ is continuous

Question

Let $f$ be a function $f:\mathbb{R}\to \mathbb{R}$ and $f(x)=x^2.$ Prove that $f$ is continuous on all of $\mathbb{R}$.

I tried this:

Let $\epsilon >0$ and assume $\mid x-x_0\mid < \delta$ for some $\delta$ and $\forall x_0 \in \mathbb{R}.$ Then $$\mid f(x)-f(x_0) \mid=\mid x^2-x_0^2 \mid=\mid x-x_0\mid \mid x+x_0\mid <\delta \mid x+x_0\mid$$

Since the last term depends on $\delta,$ I can reduce it to any $\epsilon.$ Hence, $f$ is continuous on $\mathbb{R}$.

Does this look right? I think it might fail because I don't think I can choose any old $\delta.$ Is there a way to find an exact $\delta$ that guarantee $\mid f(x)-f(x_0) \mid < \epsilon$?

A related problem. – Mhenni Benghorbal Dec 11 '13 at 03:44 — Mhenni Benghorbal, Dec 11 '13 at 03:44

score 2 · Accepted Answer · answered Dec 11 '13 at 03:29

2

Your proof is 90% of the way there. You should also deal with the $|x+x_0|$. To do that, you can restrict $\delta \leq 1$, so $|x+x_0|\leq|2x_0|+1<|2x_0|+2$ (notice these are constant terms).

Now, you can choose a $\delta$ for any $\epsilon$: $$ \delta = \min\left(1,\frac{\epsilon}{|2x_0|+2}\right) $$

answered Dec 11 '13 at 03:29

Alec

1,205

Can you explain this step: $|x+x_0|\leq|2x_0|+1$? – Dec 11 '13 at 03:47
1

$|x+x_0|\leq |x|+|x_0| \leq |x_0| + 1 + |x_0| = |2x_0| + 1$ – Alec Dec 11 '13 at 03:50

Use epsilon-delta definition to show $f(x)=x^2$ is continuous

1 Answers1