1. ## Continuity proof

I need to show that x^2 is continuous on its domain using the epsilon-delta definition. So far, I've started with |(x^2) - (c^2)|< E, which is the same as |x+c||x-c|< E. However, I'm not quite how to get from here to |x-c|< D, where D is delta (which I'm pretty sure is supposed to depend on epsilon and c).

2. Originally Posted by spoon737
I need to show that x^2 is continuous on its domain using the epsilon-delta definition. So far, I've started with |(x^2) - (c^2)|< E, which is the same as |x+c||x-c|< E. However, I'm not quite how to get from here to |x-c|< D, where D is delta (which I'm pretty sure is supposed to depend on epsilon and c).
Let $x_0 \in \mathbb{R}$ we want to show $f(x) = x^2$ is continous at that point. We want to show $|f(x) - f(x_0)|< \epsilon$ can be made small, thus, $|x^2 - x_0^2| = |x - x_0||x+x_0|< \epsilon$. Now say $|x-x_0|<\delta$ then $|x| = |x-x_0+x_0|\leq |x-x_0| + |x_0| \implies |x|+|x_0| \leq |x-x_0|+2|x_0|$ but then $|x+x_0|\leq |x|+|x_0|\leq |x-x_0|+2|x_0| < \delta + 2|x_0|$. Thus, $|f(x)-f(x_0)| = |x-x_0||x+x_0| < \delta (\delta + 2|x_0|)$ choose $\delta \leq 1$ then $|\delta (\delta + 2|x_0|) \leq \delta (1+2|x_0|)$. Thus, we have that $|f(x)-f(x_0|< \delta (1+2|x_0|)$, this means we have to chose $\delta = \min \left\{ \frac{\epsilon}{1+2|x_0|}, 1\right\}$

3. Originally Posted by kalagota

so, if $\varepsilon > 0$, choose $\delta = \frac{\varepsilon}{1+2c}$
Careful what if $c=-\frac{1}{2}$

4. Originally Posted by ThePerfectHacker
Careful what if $c=-\frac{1}{2}$

yeah, i should have edited it before i posted it.. anyways, you already had it.. thanks..

hey, can you help me with my problem i posted in the algeb section? thanks.