Proving that the limit of a function doesn't exist at an essential singularity 
Assume $f$ is analytic on a deleted neighborhood $B'(0;a)$. Prove that
  the limit of the function as $z$ approaches $0$ exists (possibly
  infinite) if, and only if there exists an integer $n$ and a function $g(z)$
  analytic on $B(0;a)$, with $g(0)$ not zero such that $f(z)=z^ng(z)$ in
  $B'(0;a)$

The "if" direction is simple but I am struggling with the other direction. The "only if" part is the same as proving that if the principle part of Laurent series expansion for the function around $0$ is an infinite sum, the limit doesn't exist which is the same as proving that it depends on the direction in which we approach $0$. I tried to exhibit two different limits but failed mainly due to the fact that I couldn't deal with the arbitrary coefficients of the series (since this should hold for any function with an essential singularity). This is a problem in Apostol's mathematical analysis and he doesn't mention the Casorati–Weierstrass theorem so I suppose one should be able to solve the problem without depending on it.

If the limit $\lim_{z\to0}f(z)$ exists in $\mathbb C$, then $a$ is a removable singularity and you can apply Riemann's theorem on removable singularities.
Otherwise, $\lim_{z\to0}f(z)=\infty$. So, there is some $r\in(0,a)$ such that $f$ has no zeros in $B(0;r)$. Let$$\begin{array}{rccc}\varphi\colon&B(0;r)&\longrightarrow&\mathbb C\\&z&\mapsto&\begin{cases}\frac1{f(z)}&\text{ if }z\neq0\\0&\text{ if }z=0.\end{cases}\end{array}$$Then $\varphi$ is analytic, $0$ is a removable singularity of $\varphi$ and you can apply Riemann's theorem to it.
