Prove that the harmonic function is identically zero Let $u$ be a harmonic function in $D(0,1)\backslash \{0\}$. Prove that if
$$\lim_{|z|\to 1} u(z)=\lim_{z\to0}\frac{u(z)}{\log|z|}=0,$$
then $u$ is identically zero on $D(0,1)$.
Here the complex number $z$ is identified with the point $(x,y)$ in the Euclidean plane, where $x=\mathrm{Re}(z),\; y=\mathrm{Im}(z)$.
Here I cannot let $u=\mathrm{Re}(f)$ for some holomorphic function $f$ since the domain $D(0,1)\backslash \{0\}$ is not simply connected. What kind of approach should I take here? Any hints or advices will help a lot!

Wlog $u$ is real-valued. You could show that $$\epsilon\ln|z|\le u(z)\le-\epsilon\ln|z|$$for every $\epsilon>0$:
Let $v(z)=u(z)+\epsilon\ln(|z|)$. If $|\xi|=1$ or $\xi=0$ the hypothesis implies that $$\limsup_{z\to\xi}v(z)\le0.$$So, since $v$ is harmonic, $v\le 0$, which says that $u(z)\le-\epsilon\ln|z|$. Similarly for the other inequality.
