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Thread: epsilon-delta definition of continuity

  1. #1
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    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) $\displaystyle f(x)=|x|$
    (ii) $\displaystyle f(x)=\sqrt{x}$
    (iii)$\displaystyle f(x)=x^3$
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TXGirl View Post
    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)$\displaystyle f(x)=x^3$
    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
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  3. #3
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    case 2?

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

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