Proving origin to be removable singularity(Proof verification)

From $$lim_{zto 0}zf(z)=0,$$ we know than $z=0$ is removable singularity of $F(z):=zf(z)$. Define $F(0)=0$, then $F$ is analytic at $z=0$, and its Taylor series of $F$ at $z=0$ is $$F(z)=zf(z)=sum_{n=1}^{infty}a_nz^n,$$ ($a_0=0$ is due to $F(0)=0$). So the Taylor series of $f$ at $z=0$ is $$f(z)=sum_{n=1}^{infty}a_nz^{n-1},$$ hence, $$lim_{zto 0}f(z)=a_1.$$ So the origin is removable singularity of $f(z)$.

A counterexample for your proof would be $f(z)=1/sqrt{z}$. I think you need to use the property of analyticity in some way to exclude such cases.