Say that $f(z) = Σ a_n z^n$ has radius of convergence 1. Can there be a complex $u$ with $|u|= 1$ and two sequences $(x_n)$ and $(y_n)$ with $|x_n| < 1$, $|y_n| < 1$, $x_n → u$, and $y_n → u$, whose limits are finite: $-∞<\lim_{x_n→u} f(x_n)<∞$, and $-∞<\lim_{y_n→u} f(y_n)<∞$, yet $\lim_{x_n→u} f(x_n) \neq \lim_{y_n→u} f(y_n)$?
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sure; for example, $f(z)=\exp \frac{z+1}{z-1}$ is bounded on the unit disc and of modulus $1$ on the unit circle except at $1$ where it is undefined and it is not hard to show that one can find $x_n \to 1, f(x_n) \to a$ for any $|a| \le 1$ – Conrad Sep 13 '21 at 19:38
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@Conrad, your example answers my question as I wrote it. I mean to ask a slightly different question. I know of the Great Pickard theorem and your example is compatible with that theorem. Put your comment in the form of an answer and I will accept it. I will enhance the question in another post. – SolutionExists Sep 14 '21 at 02:52
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no problem - done; – Conrad Sep 14 '21 at 03:06
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1For how deep your question can go, see the google search "cluster set" + boundary + behavior + analytic and my answer to Name for multi-valued analogue of a limit. Related is Andrés E. Caicedo's answer to Behaviour of power series on their circle of convergence and my answer to Power series which diverges precisely at the roots of unity, converges elsewhere. – Dave L. Renfro Sep 14 '21 at 05:32
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As a simple example $f(z)=\exp \frac{z+1}{z-1}$ is bounded on the unit disc and of modulus $1$ on the unit circle except at $1$ where it is undefined and it is not hard to show that one can find $x_n \to 1, f(x_n) \to a$ for any $|a| \le 1$
Conrad
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