# X^2 is continuous question

1. Nov 25, 2013

### tmbrwlf730

Hi everyone. So the delta-epsilon proof to show that x2 is continuous goes a little like: |f(x) - f(xo)| = |x2 - xo2| = |x - xo| |x + xo|.

Here you want to bound the term |x + xo| = |x| + |xo| by taking |x| = |x - xo + xo| = |x - xo| + |xo|.

Here you're suppose to take δ = 1 while |x - xo| < δ, so |x - xo| + |xo| < 1 + |xo|.

Putting it back into the earlier equation to get:
(1 + 2|xo|) |x - xo| < ε.

My question is why don't you set the last |x - xo| in |x + xo| |x - xo| to 1 to get (1 + 2|xo|) * 1 < ε? Why do you only set |x - xo| to 1 for the |x + xo| term but not for the other?

Thank you.

2. Nov 26, 2013

### pasmith

We don't set $|x - x_0| = 1$; we impose the bound $|x - x_0| < \delta$ and make the assumption $\delta < 1$.

Your purpose is to obtain a bound on $\delta$ in terms of $\epsilon$ (and $x_0$). Eliminating all the deltas using the assumption $\delta < 1$ defeats this purpose.

3. Nov 26, 2013

### vela

Staff Emeritus
I hope you don't really think that |a+b| = |a| + |b|.

If you were to replace $\lvert x-x_0\rvert$ in both instances by 1, you'd be showing that if $\lvert x - x_0 \rvert < 1$, then
$$\lvert f(x)-f(x_0) \rvert \lt 1 + 2\lvert x_0 \rvert.$$ That statement is fine by itself, but it's not useful in the context of the proof. Besides the reason pasmith already pointed out, there's another problem. You need to show that for any $\varepsilon > 0$, you can find a $\delta$ that works, but you can't make the claim that $1+2\lvert x_0 \rvert < \varepsilon$ for any $\varepsilon > 0$. If $x_0=1$, for example, the inequality wouldn't hold if $\varepsilon=1$. On the other hand, when you have
$$\lvert f(x)-f(x_0) \rvert \lt (1 + 2\lvert x_0 \rvert)\lvert x-x_0 \rvert,$$ you can satisfy the inequality
$$\lvert f(x)-f(x_0) \rvert \lt (1 + 2\lvert x_0 \rvert)\lvert x-x_0 \rvert < \varepsilon$$ for any $\varepsilon>0$ provided $\lvert x-x_0 \rvert$ is small enough.