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Math Help - Use of definitions to prove limits

  1. #1
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    Use of definitions to prove limits

    How would you prove the following limit using the epsilon-delta definition of a limit,

    \lim_{x\to1} \frac{x}{1+x} = \frac{1}{2}
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    How would you prove the following limit using the epsilon-delta definition of a limit,

    \lim_{x\to1} \frac{x}{1+x} = \frac{1}{2}
    Where are you stuck?
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Where are you stuck?
    Sorry, here is what I have so far
     |\frac{x}{1+x} - \frac{1}{2}| = |\frac{2x-x-1}{2x+2}| = |\frac{x-1}{2x+2}| = |\frac{1}{2x+2}||x-1|

    And we know |x-1| \leq \delta .....

    That's all I have right now. I can't figure out where to go from there
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    Sorry, here is what I have so far
     |\frac{x}{1+x} - \frac{1}{2}| = |\frac{2x-x-1}{2x+2}| = |\frac{x-1}{2x+2}| = |\frac{1}{2x+2}||x-1|

    And we know |x-1| \leq \delta .....

    That's all I have right now. I can't figure out where to go from there
    Fix \delta =1 so we can find an upper bound for \frac{1}{\left|x+1\right|}:

    \left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<3\implies\frac{1}{3}<\frac{1}{x+1}<1.

    Therefore \frac{1}{\left|2x+2\right|}\left|x-1\right|=\frac{1}{2\left|x+1\right|}\left|x-1\right|<\frac{1}{2}\left|x-1\right|<\varepsilon. Therefore, we pick \delta=\min\{2\varepsilon,1\}

    Does this make sense?
    Last edited by Chris L T521; February 7th 2010 at 07:21 PM.
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    Fix \delta =1 so we can find an upper bound for \frac{1}{\left|x+1\right|}:

    \left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2\implies\frac{1}{2}<\frac{1}{x+1}<1.

    Therefore \frac{1}{\left|2x+2\right|}\left|x-1\right|=\frac{1}{2\left|x+1\right|}\left|x-1\right|<\frac{1}{2}\left|x-1\right|<\varepsilon. Therefore, we pick \delta=\min\{2\varepsilon,1\}

    Does this make sense?
    Ok yea the only thing that never makes sense to me is this step
    \left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2

    how do you go from -1<x-1<1 to 1<x+1<2
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    Ok yea the only thing that never makes sense to me is this step
    \left|x-1\right|<1\implies -1<x-1<1\implies 1<x+1<2

    how do you go from -1<x-1<1 to 1<x+1<2
    Whoops...that should be 1<x+1<3, but even then, the upper bound on \frac{1}{\left|x+1\right|} remains the same.

    Sorry about that!
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  7. #7
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    Right, I just wanted to understand what was going on. It all makes sense now, thank you so much for the help
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