I want to proof the following claim
Let $a \in \mathbb{C}$. Consider a function $f$, where $f$ is holomorphic on $\mathbb{C}\setminus\{a\}$ and $a$ being an essential singularity of $f$. Moreover let $g$ be an entire function such that $g(a) \neq 0$. Then $h = f\cdot g$ has an essential singuarity at $z = a$.
My attempt:
$f$ has an essential singularity if $\lim_{z \to a} |(z-a)^k\cdot f(z)| = +\infty$ does not exist for every $k \in \mathbb{N}$. Otherwise we would have a pole of order $k$, if $k$ is the smallest number such that the limit exists.
Then \begin{align} \lim_{z \to a} |(z-a)^k \cdot h(z)| &= \lim_{z \to a} |(z-a)^k \cdot f(z) \cdot g(z)|\\ & = \lim_{z \to a} |(z-a)^k \cdot f(z)| \cdot \lim_{z \to a} |g(z)| \\ &= \underbrace{\lim_{z \to a} |(z-a)^k \cdot f(z)|}_{\infty} \cdot \underbrace{|g(a)|}_{\neq 0} \\ &= +\infty \end{align} for every $k \in \mathbb{N}$. Hence, $h$ has to have an essential singularity at $z = a$.
I am unsure about whether that is enough to proof the claim or not. Can someone confirm my attempt or give a hint how the statement can be proven?
Edit:
I am particulary unsure about the second step where I write
\begin{align} \lim_{z \to a} |(z-a)^k \cdot f(z) \cdot g(z)| = \lim_{z \to a} |(z-a)^k \cdot f(z)| \cdot \lim_{z \to a} |g(z)|. \end{align} Am I allowed to do this here?
Your proof is fine because $\lim_{z \to a} |g(z)| = |g(a)| \ne 0$. It might be simpler to argue in the opposite direction. If $h$ has a non-essential singularity at $z=a$ then $$ c = \lim_{z \to a} (z-a)^k \cdot h(z) $$ exists for some non-negative integer $k$. It follows that $$ (z-a)^k \cdot f(z) = \frac{(z-a)^k \cdot h(z)}{g(z)} \to \frac{c}{g(a)} $$ for $z \to a$, so that $f$ has a non-essential singularity at $z=a$, contrary to the assumption.