If $\lim_{n \to \infty} g(n) = 0$, then what can we say about $\lim_{h \to 0} \lim_{n \to \infty} g(nh)$? Let’s say I have a continuous function $g:\mathbb{R} \to \mathbb{R}$, such that : $\lim_{n \to \infty} g(n) = 0$ ($n$ is an integer).
Then I was wondering, what can I say about the following double limit :
$$\lim_{h \to 0} \lim_{n \to \infty} g(nh)$$
The problem here is that it seems that it’s $0$, because we first apply $\lim_{n \to \infty}$, but does it mean that if we switch the limits the result is $g(0)$ ?
 A: False: let $g(x)=\sin (\pi x)$. Then $\lim g(n) =0$ but $\lim g(nh) $ does not even exist for $h=\frac 1 k$, $k=1,2,...$. If $g$ is decreasing and you take limit as $h \to 0$ through positive values then the iterated limit equals the infimum of $g$ which is $0$ so the answer is 'yes'. 
A: The outer $\lim_{h\to0}$ takes any small $h$ (as in, close to $0$ but not actually $0$) and evaluates whatever is inside that limit expression. For any small, but positive $h$, the expression does evaluate to $0$, so the entire limit is also $0$.
Almost.
There are a few subtleties. One of them is that we don't know what happens for negative $h$, as a negative $h$ would mean we care about $\lim_{n\to-\infty}g(n)$. That can be solved by restricting ourselves to positive $h$ by writing $\lim_{h\to 0^+}$ instead.
The other subtlety is that when you use $\lim_{n\to\infty}$, using the letter $n$, it's implied that you're looking at integers, by convention. That doesn't have to be true. While unusual, there is nothing wrong with letting $n$ represent real numbers rather than integers. If $n$ does indeed refer only to integers, then $\lim_{n\to\infty}g(n)=0$ tells us next to nothing about $\lim_{n\to\infty}g(hn)$ as $h\to0$.
If $n$ represents integers, and we know $g$ is decreasing (or just eventually decreasing), then $\lim_{n\to\infty}g(n)=0$ implies $\lim_{x\to\infty}g(x)=0$ for real $x$, and the argument I use in the first paragraph holds completely.
