I asked a similar question here but only received a response within the context of real analysis. Since I am mainly interested in the context of complex analysis, I am posting a modified version here.
Suppose we have a function $f$ that is analytic in some region $R$, and the point $z=0$ may be on the boundary $\partial R$. It seems intuitive to me that if $\lim_{z \to 0} f(z)$ exists, then $\lim_{z \to 0} z f'(z) = 0$.
Here is an argument for this that would convince a typical physicist like myself. $f(z)$ cannot have an essential singularity at $0$, because $\lim_{z \to 0} f(z)$ would not exist. So as $z \to 0$, it scales like $f(z) = c + O( z^\alpha )$ for some constants $c$ and $\alpha$. Moreover, $\alpha > 0$ since the limit exists. Therefore, $z f'(z) = O( z^\alpha ) \to 0$.
Is this true? If so, is there a simpler or more elegant proof? If not, what additional assumptions would be needed?
Edit: The case I am most interested in is where $R$ satisfies the necessary conditions for the singularity classification theorem (see, for example, the Wikipedia page for an essential singularity). That is, assume that for every open neighborhood $N$ of $z=0$, $R \cap N$ is non-empty.
Edit: Sangchul Lee's response shows that this is not sufficient, but as I noted in my comment to his answer, intuitively, the counterexample suffers from defining the function on a very artificial domain that requires $z=0$ to be approached basically along the $x$-axis. If the function is analytically continued to a natural domain of some kind (so that essential singularities can be seen for what they are), is the result true?
We can adapt one of the counter-examples in real case to this question.
Let $R = \{ x+iy : x > 0 \text{ and } |y| < x^2 \} $ and $f(z) = z \sin(1/z)$. If $z = x+iy \in R$, then
$$ \left| \operatorname{Im}\left(\frac{1}{z}\right) \right| = \frac{\lvert y \rvert}{x^2+y^2} < \frac{x^2}{x^2+y^2} < 1. $$
This proves that we have $|\sin(1/z)| \leq C$ on $R$ for some constant $C > 0$. So $f(z) \to 0$. But
$$ z f'(z) = f(z) - \cos\left(\frac{1}{z}\right) $$
and therefore $zf'(z)$ does not converge as $z\to 0$ in $R$.