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Thread: Continuous with limit

  1. #1
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    Continuous with limit

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

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

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

    Thanks.
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    If $\displaystyle 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" $\displaystyle \mathbb R^3 $)
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    Quote Originally Posted by tttcomrader View Post
    For $\displaystyle | z | \neq 1 $, show that the following limit exists:

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

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

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

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

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

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