Can someone show how to use the epsilon-delta def of continuity to prove that the following are continuous:
(i) $\displaystyle f(x)=|x|$
(ii) $\displaystyle f(x)=\sqrt{x}$
(iii)$\displaystyle f(x)=x^3$
So to show that $\displaystyle f(x)=\sqrt{x}$ is continuous, using $\displaystyle \epsilon-\delta$ definition, I say:
Fix $\displaystyle c$ and $\displaystyle \epsilon > 0$
Case 1) Suppose $\displaystyle x>c$ and $\displaystyle |x-c|<\delta$
Choose $\displaystyle \delta=\epsilon ^2 + 2\epsilon \sqrt{c} > 0$
Then $\displaystyle x < c< 2 \epsilon \sqrt{c} + \epsilon ^2 = (\sqrt{c}+\epsilon)^2$
So $\displaystyle sqrt{x} < sqrt{c} + \epsilon$ and $\displaystyle |\sqrt{c} - \sqrt{x}| = \sqrt{x} - \sqrt{c} < \epsilon$
Case 2) Suppose $\displaystyle x<c$ and $\displaystyle |x-c|<\delta$
Can somebody tell me how to go about case 2?
We will treat the case $\displaystyle c\not =0$ (why do we have to do this?)
$\displaystyle |f(x) - f(c)| = |\sqrt{x} -\sqrt{c}|\cdot \frac{\sqrt{x} + \sqrt{c}}{\sqrt{x}+\sqrt{c}} = \frac{|x-c|}{\sqrt{x}+\sqrt{c}} \leq \frac{|x-c|}{\sqrt{c}} < \frac{\delta}{\sqrt{c}}$