Limit involving floor function: $\lim\limits_{x\to 0} x \left\lfloor\frac1x \right\rfloor$ [duplicate]

I'm studying calculus right now but I'm stuck at solving a limit involving floor function.

The problem is to find $$\lim_{x\to 0}x \left\lfloor\frac{1}{x} \right\rfloor$$ where $\lfloor\cdot\rfloor$ denotes the floor function.

My first thought was to let $x=1/t$ so when ${x\to 0+}$ then ${t\to \infty}$ so it seems $\lim_{t\to \infty}[t]/t$ doesn't exist. But I can't go any further and don't know whether my thought is correct. It seems $t=N+\delta$ doesn't help because t goes to infinity. Can it be proved by the epsilon-delta method or something else? Thank you for your help.

Solution 1:

$\frac {t-1} t \leq \frac {[t]} t \leq \frac t t$ so $\lim_{t \to \infty} \frac {[t]} t =1$. So $\lim_{x \to 0+} x[\frac 1 x]=1$. But $\lim_{t \to -\infty} \frac {[t]} t$ is also $1$ so $\lim_{x \to 0} x[\frac 1 x]=1$.

Solution 2:

We have that for any $\frac1x\in[n,n+1) \implies \left[\frac1x\right]=n$ and $x \in \left(\frac1{n+1},\frac1n\right]$ then

$$\frac{n}{n+1}\le x \cdot \left[\frac1x\right]\le \frac{n}{n}=1$$

then since $x\to 0^+ \implies n\to \infty$, by squeeze theorem the result follows.

Solution 3:

Note that for $x\ne0$, $$ \frac1x-1\lt\left\lfloor\frac1x\right\rfloor\le\frac1x $$ For $x\gt0$, $$ \overbrace{1-x}^{1-|x|}\lt x\left\lfloor\frac1x\right\rfloor\le1 $$ For $x\lt0$, $$ 1\le x\left\lfloor\frac1x\right\rfloor\lt\overbrace{1-x}^{1+|x|} $$ Therefore, for all $x\ne0$, $$ 1-|x|\lt x\left\lfloor\frac1x\right\rfloor\lt1+|x| $$ Now apply the Squeeze Theorem.

Solution 4:

$$\lim_{z\to\pm\infty}\frac{\lfloor z\rfloor}{z}=1$$ seems obvious.

For the skepticals,

$$\lim_{z\to\pm\infty}\frac{\lfloor z\rfloor}{z}=\lim_{z\to\pm\infty}\frac{z}{z}-\lim_{z\to\pm\infty}\frac{\{z\}}{z}=1-0.$$