A question on limits of functions of a sequence and removable singularities Let $f$ be analytic in $D_r(z_0)$\ {$z_0$}. Suppose for each sequence {$z_n$} in $D_r(z_0)$\ {$z_0$} such that $z_n\rightarrow z_0$ there is a subsequence {$z_{r_n}$} such that $f(z_{r_n})\rightarrow 0$. This is what confuses me. Clearly the sebsequence too converges to $z_0$. So doesn't it imply that $f(z_n)\rightarrow 0$ as well? Which then would imply that $f(z)\rightarrow 0$ as $z\rightarrow z_0$ making $z_0$ a removable singularity. Is my thinking correct? Thanks
 A: Claim: $f$ is bounded in a near  $z_0.$
If not, then given any $n \in \mathbb{N},$ there exists $z_n \in D(z_0, r)$ with $|z_n - z_0| < \frac{1}{n}$ such that $|f(z_n)| > n.$ This contradicts the given condition.
Since $f$ is bounded near $z_0,$ it has a removable singularity at $z_0.$ (to see this use use the Laurent series expression of $f$.)
$\bf{EDIT:}$ Let $f(z) = \sum_{n \in \mathbb{Z}}a_n(z-z_0)^n$ be a Laurent series expression of $f$ centered at $z_0.$ Suppose $|f(z)| \leq M$ near $z_0$ and $r > 0$ be small. Then $|a_n| \leq \frac{M}{r^n}.$ If $n < 0,$ then $\frac{M}{r^n} \rightarrow 0$ as $r \rightarrow 0.$ This shows that $a_n = 0$ for $n < 0.$ So $f$ has a removable singularity at $z_0.$ 
A: Your conclusion is correct, but your logic in getting there isn't clear.  For a given sequence $\{z_n\}$ converging to $z_0$, the fact that it has a subsequence $\{z_{r_n}\}$ such that $\{f(z_{r_n})\}$ converges to $0$ does not imply that $\{f(z_n)\}$ converges to $0$. For example, consider $f(z) =\sin(1/z)$, $z_0=0$, and $z_n=\dfrac{2}{n\pi}$. In that case $\{f(z_n)\}$ does not converge, but $\{f(z_{2n})\}$ converges to $0$.
However, the fact that such a subsequence exists for every sequence converging to $z_0$ does imply that $\lim\limits_{z\to z_0}f(z) = 0$.  (The previous example does not satisfy the hypothesis.)  To see this by contraposition, suppose that it is not the case that $f(z)\to 0$ as $z\to z_0$.  Then there exists $\varepsilon>0$ such that for all $\delta>0$ there exists $z$ with $|z-z_0|<\delta$ and $|f(z)|\geq\varepsilon$.  Thus, for some fixed $\varepsilon>0$, there exists for each $n$ a $z_n\in\mathbb C$ such that $|z_n-z_0|<\frac1n$ and $|f(z)|\geq \varepsilon$.  This implies that $\{z_n\}$ converges to $z_0$, but there exists no subsequence $\{z_{r_n}\}$ such that $\{f(z_{r_n})\}$ converges to $0$.
