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Math Help - elipson-delta definition of a limit to justify the answer.

  1. #1
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    Lightbulb elipson-delta definition of a limit to justify the answer.

    So Far I found the limit:

    Lim x->0 1/(x+1)=1

    then this is the scratchwork:
    1/(x+1)-1<E when 0< |x-0|< delta

    Not sure if I did this inequality right:

    |1/(x+1)-1|<E <==>|x/(x+1)|<E <==> |x/(x+1)|< 0E??

    Do I have to find a number M??
    1/|x+1| less than or equal to M

    Proof:
    |1/(x+1)-1|=|x|/|x+1| then I don't know how to keep going, not sure if it's 0E is correct
    Last edited by mr fantastic; September 23rd 2011 at 02:00 PM. Reason: Title.
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  2. #2
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    Re: elipson-delta definition of a limit to justify the answer Please help not sure ab

    Quote Originally Posted by LAPOSH42 View Post
    So Far I found the limit:

    Lim x->0 1/(x+1)=1

    then this is the scratchwork:
    1/(x+1)-1<E when 0< |x-0|< delta

    Not sure if I did this inequality right:

    |1/(x+1)-1|<E <==>|x/(x+1)|<E <==> |x/(x+1)|< 0E??

    Do I have to find a number M??
    1/|x+1| less than or equal to M

    Proof:
    |1/(x+1)-1|=|x|/|x+1| then I don't know how to keep going, not sure if it's 0E is correct
    If \displaystyle 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon, then \displaystyle \lim_{x \to c}f(x) = L.

    In this case, you want to show \displaystyle \lim_{x \to 0}\frac{1}{x + 1} = 1, so you need to show that \displaystyle 0 < |x - 0| < \delta \implies \left|\frac{1}{x + 1} - 1\right| < \epsilon.

    Some scratch-work...

    \displaystyle \begin{align*} \left|\frac{1}{x + 1} - 1 \right| &< \epsilon \\ \left|\frac{x}{x + 1}\right| &< \epsilon  \\ \frac{|x|}{|x + 1|} &< \epsilon \\ |x| &< |x + 1|\epsilon \end{align*}

    Now suppose we restrict \displaystyle |x| < \frac{1}{2} (say), then \displaystyle -\frac{1}{2} < x < \frac{1}{2} \implies \frac{1}{2} < x + 1 < \frac{3}{2} \implies \frac{1}{2} < |x + 1|, therefore

    \displaystyle \begin{align*} |x| &< |x + 1|\epsilon \\ |x| &< \frac{1}{2}\epsilon \end{align*}

    So if we let \displaystyle \delta = \min\left\{\frac{1}{2}, \frac{1}{2}\epsilon\right\} and reverse the process, you will have your proof.
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    MHF Contributor Siron's Avatar
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    Re: elipson-delta definition of a limit to justify the answer Please help not sure ab

    If you want to have more examples of proves using \epsilon-\delta definition then I recommand you to take a look here:
    http://www.mathhelpforum.com/math-he...ofs-47767.html
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    Question Re: elipson-delta definition of a limit to justify the answer

    Proof: let E>0 and define delta= min {1/2,1/2E}
    then if 0<|x-0|<delta

    |1/(x+1)-1| = |x/x+1|
    =|x|/|x+1|
    <(1/2)?|x-0|
    <(1/2)?2E
    =E
    Last edited by LAPOSH42; September 23rd 2011 at 02:17 PM.
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