I'm asked to evaluate, for $x \to 0$ and $x \to +\infty$, the limit:
$$\lim xM \left(\frac{1}{x^4} \right)$$
where $M:=x-\lfloor x \rfloor$. I've also been told that one of them does not exist, and I should demonstrate that.
I plotted the functions ($M(x)$ is in orange, the main one is in blue):
I think that:
- $\displaystyle\lim_{x \to \infty} xM\left(\frac{1}{x^4}\right) = 0$, from the graph, but I'm not able to evaluate it algebraically. I was thinking that $M(x)$ is not continuous in $0$, but it is from the right. And since $\frac 1{x^4} \to 0^+$ we could have that $M\left(\frac 1 {x^4}\right) \to 0$. But then $x \to +\infty$ and I'm not able to proceed.
- $\displaystyle\lim_{x \to 0} xM\left(\frac 1 {x^4}\right) = 0$. Since $M(x) \in [0, 1)$, so by the squeeze theorem the limit is $0$ (we have that $0 < xM\left(\frac 1 {x^4}\right) \le x$).
Since one of them should not exist (so I'm told) I think there's an error in my reasoning.