Assume $f$ is defined on some domain $D\subset\mathbb C$, differentiable on $D$ as a function of two real variables, and that, for all $z\in D$, the limit $\lim\limits_{h \to 0} \Re\left(\frac{f(z+h)-f(z)}{h}\right)$ exists and is finite. Does this imply that $f$ is holomorphic on $D$?
This was a task on my exam today, which I didn't manage.
Let us fix some $z$ in $D$ and use the notations $$h=r+is\qquad f_x(z)=a+ib\qquad f_y(z)=c+id$$ for some real numbers $r$, $s$, $a$, $b$, $c$ and $d$. Then the $\mathbb R^2$-differentiability of the function $f$ at $z$ reads, in the limit $h\to0$ or, equivalently, in the limit $(r,s)\to(0,0)$, $$f(z+h)-f(z)=(a+ib)r+(c+id)s+o(|h|)$$ that is, $$\frac{f(z+h)-f(z)}h=\frac{(f(z+h)-f(z))\bar h}{|h|^2}=\frac{((a+ib)r+(c+id)s)(r-is)}{r^2+s^2}+o(1)$$ Thus, the hypothesis is exactly the existence of the limit $$\lim_{(r,s)\to(0,0)}\frac{ar^2+(b+c)rs+ds^2}{r^2+s^2}$$ Considering the limits of this ratio when $r\to0$ along the lines $s=rt$, for each $t$ fixed, one sees that $$\frac{a+(b+c)t+dt^2}{1+t^2}$$ should be independent of $t$. In particular, the cases $t=0$ and $t\to\infty$ yield the condition $$a=d$$ and the cases $t=\pm1$ yield the condition $$b+c=0$$ Since these are Cauchy-Riemann equations for $f$ at $z$, the proof is complete.