positive harmonic function has a zero limit at a point on the boundary Let $u$ be a positive harmonic function in $\{ \Re{z} > 0\}$ such that 
$\lim_{r \rightarrow 0^+} u(r) = 0 $.
Prove that then $\lim_{r \rightarrow 0^+} u(re^{i\theta}) = 0 $ for any $\theta \in (-\pi/2, \pi/2)$.
What I have tried is to consider the equivalent problem by transforming the domain to the unit disc using the Moebius map $z \rightarrow \frac{z-1}{z+1}$. Then we have 
$\lim_{r \rightarrow -1^+} u(r) = 0 $ and we need to prove that this limit is zero when approaching along the arcs of other circles which intersect the unit circle perpendicularly at $-1$ or in other words the limit is the same when approaching from different non-tangential directions. Note also we have a simply connected domain and hence a holomorphic function, $f$ exists whose real part is $u$. Also since u was positive, hence $f$ maps the unit disc to itself which means it is bounded. This in turn means that $f$ has a finite limit a.e. on the boundary. Any ideas how to proceed or how to prove this otherwise?

The assumption that $u$ is positive should bring to mind Harnack's inequality. Let's state its special case for the disk: if $u$ is positive and harmonic in $\{z:|z-z_0|<R\}$, then 
$$\frac{R-|z-z_0|}{R+|z-z_0|}u(z_0)\le u(z) \le \frac{R+|z-z_0|}{R-|z-z_0|}u(z_0)$$
for all $z$ in the disk.
The precise form of constants does  not matter; the fact that they depend only on $|z-z_0|/R$ is a consequence of the scale invariance of the Laplace equation. 
Apply the above with $z=re^{i\theta}$ and  $z_0=R=r\sec\theta$. Observe that $|z-z_0|/R=|\sin\theta|<1$, independently of $r$. Thus,
$$u(re^{i\theta})\le \frac{1+|\sin\theta|}{1-|\sin\theta|}u(r\sec\theta)$$
from where the conclusion $u(re^{i\theta})\to 0$ follows.
