Product of a monotone increasing function and a monotone decreasing function Suppose $g:\mathbb{R} \rightarrow [0,\infty)$ is a strictly increasing function such that $\lim g(x) = \infty$ as $x \rightarrow \infty$.
Suppose $h:\mathbb{R} \rightarrow [0,\infty)$ is a strictly decreasing function such that $\lim h(x) = 0$ as $x \rightarrow \infty$.
Consider the product function $f(x)=g(x)h(x)$. Is it possible to construct such $g,h$ such that $f$ does not have a limit as $x \rightarrow \infty$?
Even better, can such an $f$ be constructed if we insist that $g(x)=x$?

Take
$$ g(x) = x 2^{\lfloor x/2\rfloor}, h(x) = \frac{1}{x} 2^{-\lfloor (x+1)/2\rfloor} $$
The product oscillates between 1 and 1/2: let $f = g\cdot h$, $f(2n) = 1$, $f(2n+1) = 1/2$ for $n$ an integer. 
Edit: If you want $g(x) = x$. Take $h(x)$ to be the following (continuous) function: let $h(e^{2n}) = e^{-2n}$, and let $h(e^{2n+1}) = e^{-2n - \frac12}$, and linearly interpolate. Then $f(e^{2n}) = 1$ and $f(e^{2n+1}) =\sqrt{e} \neq 1$ for all natural numbers $n$. So $f$ does not have a limit. 
