Analytic function on upper half plane converging along two rays. Let $H=\{z\in \mathbb{C}:\Im(z)>0\}$ be the open upper half plane, and $\hat H=\bar H-\{0\}$ be the closure of $H$ minus the point $0$. Let $f$ be a holomorphic function in $H$ and bounded and continuous in $\hat H$. Show that, if $\lim_{t\rightarrow 0} f(t)=0$ for real $t$, then $\lim_{z\rightarrow 0}f(z)=0$.
For this problem, I have found two related results but not a complete solution. 
First, suppose that $f$ is instead holomorphic and bounded on an open neighbourhood of $\hat H$ that does not contain $0$. Let $\{z_k\}$ be a sequence converging to zero. Consider the sequence of functions, $\{f_k\}$ defined by $f_k(z)=f(|z_k|z)$. This sequence of functions is a normal family, hence there is a sub-sequence that converges uniformly on compact sets to a holomorphic function $g$. For any $x\in\mathbb{R}$, we have $g(x)=\lim_{k\rightarrow\infty} f(|z_k|x)=0$. So, $g(z)=0$ for all $z$. Therefore, since $f_{n_k}$ converge uniformly to $0$ on the upper semi-circle, we have that $f(z_{n_k})\rightarrow 0$. So $f(z_k)\rightarrow 0$.
The second result, is the following: If $0<\theta_1<\theta_2<\pi$, and $\{z_k\}$ is a sequence converging to zero, such that $\theta_1<\arg z_k<\theta_2$ for all $k$, then $f(z_k)\rightarrow 0$. I did this by estimating $|f(z)|$ in a neighbourhood of $0$, using the Cauchy integral formula.
 A: Your second result is on the right track. The non-tangential  assumption can be removed by considering  a  semicircle with the center at $\operatorname{Re}z_k$. 
Here's one way. Let $M$ be the maximum of $|f|$ on the unit halfdisk. Given $\epsilon>0$, find $\delta$ such that $|f|<\epsilon$ on the line segment $[-\delta,\delta]$. Fix $z$ with $|z|<\epsilon\delta $. Let $\Omega$ be the semi-circle of radius $\delta/2$ centered at $\operatorname{Re}z$. 
Somehow I don't see how to use the Cauchy estimate here, so I'll go with the maximum principle. Let  $w$ be the point with $\operatorname{Re}w=\operatorname{Re}z$ and $\operatorname{Im}w = - \delta \epsilon/M $. Pick small $\eta>0$. The function 
$$g(\zeta)=\frac{z^\eta f(\zeta)}{ \zeta-w}$$ is holomorphic in $\Omega$ and bounded by $2M/\delta$ on  $\partial \Omega$. The factor $z^\eta$   makes sure it tends to $0$ at $0$.  Hence $|g(z)|\le 2M/\delta$. Since this is independent of $\eta$, we can send $\eta\to 0$ and obtain
$$|f(z)|\le 2M|z-w|/\delta<3M\epsilon $$ or something of this kind. 
