$D= \{z: 0\lt |z-z_0|\lt r \}$, $z_0\in\Bbb C$, $r\gt0$. Let $u$ be a harmonic function in $D$, and $$\lim_{z\to z_0} \frac{u(z)}{\log |z-z_0|}=0$$ Show that $\lim\limits_{z\to z_0} u(z) $ exists.
We need an auxiliary function. Maybe the problem is related to the Dirichlet Problem
Thanks in advance for any help!
This follows easily from the general representation of harmonic functions in a punctured disc (or annulus) as $u(z)=\Re{f(z)}+a\log{|z-z_0|}$ where $f$ is analytic in the punctured disc (annulus) and $a$ complex with $2\pi a=\int_{|z-z_0|=\rho <r}{\frac{\partial u}{\partial n}}ds$.
(by Green's theorem the integral above doesn't depend on $0< \rho <r$ and $u$ has a harmonic conjugate iff the integral is zero, and the choice of $a$ makes it so for the harmonic $u-a\log{|z-z_0|}$)
But now $f$ cannot have a singularity at $z_0$ since any such (pole, essential) blows up much faster than $\log{|z-z_0|}$ so the limit in the OP would be infinite or not exist, hence $f$ extends to the disc and then obviously the limit in the OP is $a$ so we get $a=0$ and $u$ extends to a harmonic function on the disc!