I am not sure if this way of solving is correct, but here's my attempt.
For the limit to exist, we suppose f(z), a holomorphic function exists where f(z) = $\frac{1}{|z|}$ on D(0,1) \ {0}
However, f(z) = $\frac{1}{(z\bar{z})}$ where there exists zbar and hence f(z) is not holomorphic.
Thus, there does not exist any f(z).
As already said in the comments, you have only shown that a certain function $f$ does not satisfy the conditions, but you have to show that no holomorphic function $f$ with that property exists.
So assume that $f$ is such a function, and define $g$ in $\Bbb D \setminus \{ 0 \}$ by $g(z) = zf(z)$. Then $g$ is bounded in a neighborhood of $z=0$, and Riemann's theorem on removable singularities states that $g$ has a removable singularity at $z= 0$. In particular, $\lim_{z\to 0} g(z)$ exists (and is not zero).
It would follow that $$ \frac{z}{\lvert z \rvert} = \frac{g(z)}{\lvert z \rvert f(z)} $$ has a limit for $z \to 0$, which is not the case: The left-hand side is equal to $1$ for positive real $z$, and equal to $-1$ for negative real $z$. This is a contradiction.