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

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.

• A hint on $\LaTeX$: there are symbols \lfloor and \rfloor which render as $\lfloor$ and $\rfloor$ to represent the floor function. – CiaPan Oct 28 '19 at 8:11
• @CiaPan Thank you for your advice. – Scott Lee Oct 28 '19 at 8:31

$$\frac {t-1} t \leq \frac {[t]} t \leq \frac t t$$ 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 $$\lim_{x \to 0} x[\frac 1 x]=1$$.

• Thank you! I didn't expect the right limits exists. – Scott Lee Oct 28 '19 at 8:34

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.

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

For the skepticals,

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

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.