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Math Help - Complex Limit z \to i

  1. #1
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    Complex Limit z \to i

    \displaystyle\lim_{z\to i}\frac{z^2-1}{z^2+1}=\frac{(z+1)(z-1)}{(z+i)(z-i)}

    This limit is giving me a little trouble.

    I tried factoring but I wasn't able to go anywhere from there.

    Anyone have any suggestions?

    Thanks.
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  2. #2
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    May be this helps

    <br />
\displaystyle { \frac{z^2+1-2}{z^2+1}=1-\frac{2}{z^2+1}<br />
}<br />
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  3. #3
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    I am not sure how that is going to help though. I will still have 2/0.
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    I am not sure how that is going to help though. I will still have 2/0.
    Maybe it tells you the limit doesn't exist...
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  5. #5
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    It is infinity. I just don't know how to show it.
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  6. #6
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    Suppose it were \displaytype \lim_{x\to 0}\frac{1}{x}. How would you show that goes to infinity?
    For z close to i, z^2- 1 will be close to -2 and z+ i will be close to 2i so
    \frac{z^2- 1}{(z+i)(z-i)} will be close to \frac{i}{2i(z- i)}= \frac{1}{2}\frac{1}{z- i}. You want to show that, given any M, there exist r such that if |z- i|< r, then \frac{1}{2}\left|\frac{1}{z-i}\right|< M. That should be pretty straight forward.
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