if $f(k/N)\rightarrow0$ as $N\rightarrow\infty$ for any $k$, must $f(h)\rightarrow0$ as $h\rightarrow0$? If $f$ is a function defined on [$\mathbb{R}$ or $\mathbb{C}$] such that for any [real or complex] $k$, $f(\frac{k}{N})\rightarrow0$ as $N\rightarrow\infty$ in $\mathbb{N}$, must it be true that $f(h)\rightarrow0$ as $h$ tends to $0$?
(This is essentially 2 questions, but I'd be grateful for an answer to either! I put them together because they're probably both answerable in the same way, since no derivatives are involved.)
I've tried to prove the answer is yes using a delta-epsilon argument, but keep stalling because although $f(\frac{k}{N})$ must tend to zero for any particular $k$, there's no uniform rate of convergence for all $k$ in a small interval. On the other hand, I can't find a crazy counterexample either, since once $f$ has been defined at a point, that point can be part of infinitely many sequences $\frac{k}{N}$, along all of which $f$ tends to zero.
Many thanks for any help with this!
 A: Let $f$ be a discontinuous solution of the Cauchy functional equation $f(x+y)=f(x)+f(y)$. Then for real $k$, the limit of the OP, as $N\to\infty$ over the integers,  is $0$. But $f$ cannot be continuous at $0$.
Remark: I would guess that one can produce a more "concrete" example that does not require the Axiom of Choice. 
A: For the real case, let
$$f(x) = \begin{cases} n\cdot x &,\: x/\pi^n \in \mathbb{Q}\\
\;\,0 &, \text{ otherwise}
\end{cases}$$
Since $\pi$ is transcendental, $f$ is well-defined, and it satisfies
$$\lim_{N\to\infty} f\left(\frac{k}{N}\right) = 0$$
for all fixed real $k$ and rational $N$. But it's not continuous in $0$, since every non-degenerate interval contains rational multiples of every power of $\pi$.
For the complex case, we can do something more easily visualised, let
$$f(x+iy) = \begin{cases}\qquad0 &, x = 0\\ \frac{y\sqrt{x^2+y^2}}{x} &, x\neq 0\end{cases}$$
This $f$ is continuous on each line through $0$, but the slope of the line tends to $\pm\infty$ as the angle approaches $\pm \dfrac{\pi}{2}$.
