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): M(x) and xM(1/x^4)

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.

  • $\begingroup$ You can check that $M(\varepsilon)=\varepsilon$ if $0<\varepsilon<1$. $\endgroup$ – Hanul Jeon Nov 1 '14 at 10:31
  • $\begingroup$ @tetori: Yes, it follows from the fact that $M(x) \in [0, 1)$, but I don't understand how to apply it in this case. $\endgroup$ – rubik Nov 2 '14 at 11:25
  • $\begingroup$ $xM(1/x^4)=x^{-3}$ for $x>1$ so you can calculate the case $x\to \infty$. $\endgroup$ – Hanul Jeon Nov 2 '14 at 13:26
  • $\begingroup$ @tetori: Ohh yes, thanks! $\endgroup$ – rubik Nov 2 '14 at 14:41

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