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Math Help - Continuous with limit

  1. #1
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    Continuous with limit

    For  | z | \neq 1 , show that the following limit exists:

     f(z) = \lim _{n \rightarrow \infty } \frac {z^n -1 }{z^n + 1 }

    How to define f(z) when  |z| = 1 in such a way to make f continuous?

    Thanks.
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    If z \in \mathbb C then you then the range is a unit disk down at -1 on the z-axis. (supposing you work in "kind of"  \mathbb R^3 )
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    Quote Originally Posted by tttcomrader View Post
    For  | z | \neq 1 , show that the following limit exists:

     f(z) = \lim _{n \rightarrow \infty } \frac {z^n -1 }{z^n + 1 }

    How to define f(z) when  |z| = 1 in such a way to make f continuous?

    Thanks.
    Since |z|\not = 1 it means |z| > 1 or |z| < 1.

    If |z| < 1 it means z^n \to 0 because |z|^n \to 0.
    Thus, in that case \lim \frac{z^n-1}{z^n+1} = -1.

    If |z| > 1 then \frac{z^n - 1}{z^n + 1} = \frac{1 - \left( \tfrac{1}{z} \right)^n}{1 + \left( \tfrac{1}{z}\right)^n} \to 1 because \left| \frac{1}{z} \right| < 1.

    The function f cannot be extended to a continous function on \mathbb{C}. This is because f is not continous on |z| = 1. It has one value inside the disk and it has another values outside the disk. Thus, it cannot be continous.
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