We have
Propostion 14.3.3 (Intercchange of limits and uniform limits). Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, with Y complete, and let E be a subset of X. Let $(f^{(n)})^\infty_{n=1}$ be a sequence of functions from E to Y, and suppose that this sequence converges uniformly in E to some function $f:E \to Y$. Let $x_0 \in X$ be an adherent point of E, and suppose that for each n the limit $\lim_{x\to x_0; x \in E}f^{(n)}(x)$ exists. Then the limit $\lim_{x\to x_0; x \in E}f(x)$ also exists, and is equal to the limit of the sequence $(\lim_{x\to x_0; x \in E}f^{(n)}(x))^\infty_{n=1}$; in other words we have the interchange of limits
$\lim_{n\to \infty}\lim_{x\to x_0; x\in E} f^{(n)}(x) = \lim_{x\to x_0; x\in E} \lim_{n\to \infty} f^{(n)}(x)$
Below are some details which will require in my question.
(1) extended complex plane $\mathbb C\cup ${$\infty$} is complete Metric space.Where the metric we have defined using projection by sphere.denote this metric as $d$.
(2) Now let $g:\mathbb C-(0) \to \mathbb C $ defined by $g(z)=e^{\frac{1}{z}}-1$ for all z $\in \mathbb C$.clearly it has Essential singularity at $z_0=0$.and i read in book that if function $f$ has essential singularity at some point say $z_0$ then $\lim_\limits{z \to z_0} f(z) $ not exist in extended complex plane.
but using Propostion 14.3.3 somehow ,here i got $\lim_\limits{z \to 0} \left(e^{\frac{1}{z}}-1\right)=\infty $.i know it is certainly false, but i don't know where i done mistake.Here are my calculations.
Here Laurent expansion of $g$ centered at 0 is .
g(z)= $\sum^{\infty}_{n=0} a_n z^n$ + $(\frac{1}{z}+\frac{1}{z^{2}2!}+\ldots)$ with all $a_n=0$. so we are left with
$g(z)=(\frac{1}{z}+\frac{1}{z^{2}2!}+\ldots)$ for all $z\in \mathbb C-(0)$, and here series is uniformly converges in $\mathbb C-(0)$.(as it is power series).
define $f_n:\mathbb C-(0) \to \mathbb C $ by $f_n(z)=\sum^{n}_{i=1}\frac{1}{z^{i}i!}$ for $n \in \mathbb N$
clearly fn converges uniformly to g on $\mathbb C-(0)$ ,with $\lim_\limits{z \to 0} f_{n}(z)=\infty$ for all $n \in \mathbb N$, .
so here by comparing with propostion 14.3.3.we have
$E=\mathbb C-(0)$
$Y= \mathbb C\cup ${$\infty$}
$f_n \text{are same} f_n$
$f \quad is \quad g$
$x_0=0$ and $\lim_\limits{z \to 0} f_{n}(z)=\infty$ for all $n \in \mathbb N$, . so we are ready to use that proposition. so by using that proposition we have
$\lim_\limits{z \to 0} g(z)=$$\lim_\limits{n \to \infty} \lim_\limits{z \to 0} f_{n}(z)$ (after interchanging). but then as $\lim_\limits{z \to 0} f_{n}(z)=\infty$ for all $n \in \mathbb N$
we have
$\lim_\limits{z \to 0} g(z)=$$\lim_\limits{n \to \infty} \infty $(here terms of sequence are $\infty$ and i think its make sense as we have metric space with $\infty$.so we should consider it as just a element).
but then $\lim_\limits{z \to 0} g(z)=\infty$ (constant sequence).
so we have essential singularity at 0 but still i am getting limit $\infty$. so my question is what i am missing?
Let $f(z)= \sum_{n=0}^{\infty}a_nz^n$ be a power series with radius of convergence $R \in (0, \infty].$
Then the serie is pointwise convergent in $ \{z: |z|< R\}$, but in general the convergence is not uniform in $ \{z: |z|< R\}$. The power series only converges uniformly on each compact subset of $ \{z: |z|< R\}$ !
Hennce your series of $g$ only converges uniformly on each compact subset of $\mathbb C \setminus \{0\}.$