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Thread: Proving continuity at a point.

  1. #1
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    Proving continuity at a point.

    Show that f(x) = x^3 is continuous at 1 (from the definition)

    (For all epsilon >0) (There exists delta >0) (For all x E X) if |x-1| < delta this implies
    |x^3 - 1^3| < epsilon


    |x^3 - 1^3| = |x-1| |x^2 + x + 1|


    I've not got any further with this. Could I grab some help!
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  2. #2
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    so you have factored out x-a so now you want to use delta to put a bound on that, you then know what x will vary between and so you will know the max value of |x^2+x+1|, does that help?

    $\displaystyle \displaystyle\mid(f(x)-L)$

    $\displaystyle \displaystyle\mid(x^3-1)\mid=\mid(x-1)\mid\mid(x^2+x+1)\mid$

    We then use $\displaystyle \displaystyle\delta$ to bound $\displaystyle \displaystyle\mid(x-1)\mid$

    so let $\displaystyle \displaystyle\delta=1\rightarrow\mid(x-1)\mid\le1$ so we know that x varies between 0 and 2

    so $\displaystyle \displaystyle\((x^2+x+1)\le7$
    Last edited by hmmmm; Dec 3rd 2010 at 03:08 AM.
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  3. #3
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    Quote Originally Posted by hmmmm View Post
    so you have factored out x-a so now you want to use delta to put a bound on that, you then know what x will vary between and so you will know the max value of |x^2+x+1|, does that help?
    Erm, not really. I know I need to find the delta but I'm not sure what ..
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  4. #4
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    If a given delta works, then any smaller delta works as well. Thus, you may assume that $\displaystyle |x - 1|$ is less than some fixed value, let's say $\displaystyle |x-1|<1$

    Then $\displaystyle -1 < x-1 < 1$, and so $\displaystyle 0 < x < 2$.

    So, $\displaystyle x^2 + x + 1 < 2^2 + 2 + 1 = 7$, and $\displaystyle x^2 + x + 1 > 0^2 + 0 + 1 = 1$. So $\displaystyle -7 < 1 < x^2 + x + 1 < 7$, or equivalently $\displaystyle |x^2 + x + 1| < 7$
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  5. #5
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    Quote Originally Posted by hmmmm View Post
    so you have factored out x-a so now you want to use delta to put a bound on that, you then know what x will vary between and so you will know the max value of |x^2+x+1|, does that help?

    $\displaystyle \displaystyle\mid(f(x)-L)$
    $\displaystyle \displaystyle\mid\=\mid(x^3-1)\mid\=\mid(x-1)\mid\mid(x^2+x+1)\mid$

    We then use $\displaystyle \displaystyle\delta$ to bound $\displaystyle \displaystyle\mid(x-1)\mid$
    so let $\displaystyle \displaystyle\delta=1\rightarrow\mid(x-1)\mid\le1$ so we know that x varies between -1 and 2
    so $\displaystyle \displaystyle\(x^2+x+1)\le7$
    If delta = 1, and |x-1| < 1 then surely x varies between 0 and 2?
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  6. #6
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    Ok so this is my proof:

    Let |x-1| < 1 so 0<x<2

    Therefore |x^2 + x + 1| < 7

    Let delta = epsilon/7

    Suppose |x-1| < delta
    |x^3 - 1^3| = |x-1||x^2 + x + 1| < 7|x-1| < 7delta = epsilon


    Correct?
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  7. #7
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    Almost - you have to let delta = min{1, epsilon/7}.

    This is because in your argument you assumed that delta was less than 1. So if epsilon/7 turns out to be bigger than 1, then delta = epsilon/7 doesn't work.
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  8. #8
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    Quote Originally Posted by DrSteve View Post
    Almost - you have to let delta = min{1, epsilon/7}.

    This is because in your argument you assumed that delta was less than 1. So if epsilon/7 turns out to be bigger than 1, then delta = epsilon/7 doesn't work.
    Oh yeh, must be careful with that.

    Thanks!
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  9. #9
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    yeah sorry i just made a little mistake there but i was of no consequence,
    let $\displaystyle \displaystyle\epsilon$ be given choose $\displaystyle \displaystyle\delta=min\{1,\epsilon/7\}$

    We then know that if $\displaystyle \displaystyle{0}\le\mid(x-1)\mid\le\delta\le{1}$
    then
    $\displaystyle \displaystyle\mid(x^3-1)\mid\le7\mid(x-1)\mid$

    so we conclude that

    $\displaystyle \displaystyle0\le\mid(x-1)\mid\le\delta\rightarrow\mid(x^3-1)\mid\le7*\mid(x-1)\mid\le7*\frac{\epsilon}{7}=\epsilon$
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