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Thread: Epsilon/Limits Help

  1. #1
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    Epsilon/Limits Help

    I posted a question about these a few days ago and am still having trouble understanding this. If someone could show me how to do this problem with some explanation hopefully I can understand it better.

    Find a value of alpha > 0 such that for all x, 0 < abs value (x- Xo) < alpha -> a<x<b.

    a= -7/2 b= -1/2 Xo= -3

    Our teacher barely does an examples in class, and I don't even understand what this problem is looking for. Thanks for help!
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    Find a value of alpha > 0 such that for all x, 0 < abs value (x- Xo) < alpha -> a<x<b.
    a= -7/2 b= -1/2 Xo= -3
    $\displaystyle \left| {a - x_0 } \right| = \left| {\frac{{ - 7}}{2} + 3} \right| = \frac{1}{2}$
    $\displaystyle \left| {b - x_0 } \right| = \left| {\frac{{ - 1}}{2} + 3} \right| = \frac{5}{2}$
    The smaller of those two is $\displaystyle \frac{1}{2}$.
    Therefore $\displaystyle \alpha =\frac{1}{2}$.

    To see why consider:
    $\displaystyle \left| {x - x_0 } \right| < \alpha \Rightarrow \left| {x + 3} \right| < \frac{1}
    {2} \Rightarrow - \frac{1}
    {2} < x + 3 < \frac{1}
    {2} \Rightarrow \frac{{ - 7}}
    {2} < x < - \frac{5}
    {2} < - \frac{1}
    {2}
    $
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