# Thread: epsilon-delta definition of continuity

1. ## epsilon-delta definition of continuity

Can someone show how to use the epsilon-delta def of continuity to prove that the following are continuous:

(i) $f(x)=|x|$
(ii) $f(x)=\sqrt{x}$
(iii) $f(x)=x^3$

2. Originally Posted by TXGirl
Can someone show how to use the epsilon-delta def of continuity to prove that the following are continuous:

(i) $f(x)=|x|$
(ii) $f(x)=\sqrt{x}$
(iii) $f(x)=x^3$
see here

that is for |x|

TPH and I also did the proof for x^3, it's lying around the forum somewhere, do a search.

you should be able to figure out how to do the one for sqrt(x) from those

good luck

3. ## case 2?

So to show that $f(x)=\sqrt{x}$ is continuous, using $\epsilon-\delta$ definition, I say:

Fix $c$ and $\epsilon > 0$

Case 1) Suppose $x>c$ and $|x-c|<\delta$

Choose $\delta=\epsilon ^2 + 2\epsilon \sqrt{c} > 0$

Then $x < c< 2 \epsilon \sqrt{c} + \epsilon ^2 = (\sqrt{c}+\epsilon)^2$

So $sqrt{x} < sqrt{c} + \epsilon$ and $|\sqrt{c} - \sqrt{x}| = \sqrt{x} - \sqrt{c} < \epsilon$

Case 2) Suppose $x and $|x-c|<\delta$

Can somebody tell me how to go about case 2?

4. We will treat the case $c\not =0$ (why do we have to do this?)

$|f(x) - f(c)| = |\sqrt{x} -\sqrt{c}|\cdot \frac{\sqrt{x} + \sqrt{c}}{\sqrt{x}+\sqrt{c}} = \frac{|x-c|}{\sqrt{x}+\sqrt{c}} \leq \frac{|x-c|}{\sqrt{c}} < \frac{\delta}{\sqrt{c}}$