If for all $x>0$, we have $\lim_{n\to\infty} g(xz_n)=+\infty$, does this imply $\lim_{x\to\infty} g(x)=\infty$? Let $g:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ be some function (not necessarily continuous), and let $(z_n)_{n\geq 1}$ be an increasing, diverging sequence of positive real numbers with the following property:
For all $x>0$, we have $\displaystyle\lim_{n\to\infty} g(xz_n)=+\infty$.
Then is it true that $\displaystyle\lim_{x\to\infty} g(x)=\infty$?
Intuitively I would say that this isn’t true because we have only information on a very small subsample of diverging sequences, but still we have a continuum at hand which makes it hard to cook up a counter example.
 A: Your intuition is correct. Take $$g(x):=\begin{cases}0&\text{if }x=\pi^k\text{ for some }k\in\mathbb N,\\x&\text{otherwise}\end{cases}$$ and $z_n:=n$. For every $x>0$, the sequence $(nx)_{n\in\mathbb N}$ hits a power of $\pi$ at most once: Assume to the contrary that there exist natural numbers $n>m,k>l$ such that $$nx=\pi^k,mx=\pi^l\Longrightarrow\pi^{k-l}=\frac nm$$ but this is impossible since $\pi$ is transcendental. Thus, for every $x>0$, eventually we have $g(nx)=nx$ for all $n$ and thus $\lim_{n\to\infty}g(nx)=\infty$, but $g$ does not diverge to infinity.
A: Define $g:{\mathbb R}_{>0}\rightarrow {\mathbb R}_{>0}$ by $$g(x)=\left\{\begin{array}{cc}\frac 1 p&{\rm if~}x=\frac q p~{\rm and~}p>1\\
                  x&~{\rm otherwise,}\end{array}\right.$$ where the fraction is in reduced form.
Let $\{z_n\}$ be defined by $$z_n=n!,n\in {\mathbb N.}$$ Then $z_n$ is monotonically increasing, unbounded, and for every $x>0,$ one has $$\lim_{n\rightarrow \infty} g(xz_n)=\infty.$$  However $g(x)$ does not diverge to infinity.
A: It must be that either $\displaystyle\lim_{x\rightarrow\infty} g(x) = \infty$ or this limit does not exist.
Reason:  Taking $x=1$ and $z_n$ to be the known sequence of inputs that goes to infinity for which we know the sequence of outputs $g(z_n)$ has $\displaystyle\lim_{n\rightarrow\infty} g(z_n) = \infty$ guarantees that no real number can be an upper bound for the outputs of this function. (For any potential bound $B$ there is some term in the sequence $z_n$ for which $g(z_n)>B$).
