Evaluating limit associated with greatest integer function

Question

$1)$ Find $\lim_{x \to 0} x \big [\frac{1}{x}]$

I was asked the problem in an interview. I tried to think it intuitively. As $x$ tends to $0$ from right (say $x=0.001$) then $\big [\frac{1}{x}]$ will be equal to $1000$. Multiplying with $x=0.001$ it gives $1$. So, I said that though $x$ is tending to $0$(but not equal to zero) eventually $\big [\frac{1}{x}]$ will be a large number. If this large number is multiplied with a very small number it will tend to $1$.

But the interviewer was not satisfied with my explanation and pointed that I am just considering a particular case and asked if I have any fact that holds true for all $x$ tending to zero. According to them, though my answer is correct , it is not clear as a mathematical proof.

Is their any other way of doing it?

Answer 1

(1) $x>0$

If $\lfloor \frac{1}{x}\rfloor=n$, $n\le\frac{1}{x}<n+1$ and hence $\frac{1}{n+1}<x\le\frac{1}{n}$.

$$\frac{n}{n+1}<x\lfloor \frac{1}{x}\rfloor<\frac{n+1}{n}$$

As $x\to0^+$, $n\to\infty$ and $x\lfloor\frac{1}{x}\rfloor\to1$

(2) $x<0$

If $\lfloor \frac{1}{x}\rfloor=-n$, $-n\le\frac{1}{x}<-n+1$ and hence $\frac{-1}{n-1}<x\le\frac{-1}{n}$.

$$\frac{n}{n-1}>x\lfloor \frac{1}{x}\rfloor>\frac{n-1}{n}$$

As $x\to0^-$, $n\to\infty$ and $x\lfloor\frac{1}{x}\rfloor\to1$

Answer 2

For $x>0$, $\left[\dfrac{1}{x}\right]\leq\dfrac{1}{x}$, so $x\left[\dfrac{1}{x}\right]\leq 1$. Now $\dfrac{1}{x}\leq\left[\dfrac{1}{x}\right]+1$, then $1-x\leq x\left[\dfrac{1}{x}\right]$, then taking $x\rightarrow 0^{+}$, then $x\left[\dfrac{1}{x}\right]\rightarrow 1$.

Similar procedure for $x<0$: For $y<0$, then $-[y]-1=[-y]$, so for $x<0$, $x\left[\dfrac{1}{x}\right]=(-x)\cdot-\left[\dfrac{1}{x}\right]=(-x)\left(\left[-\dfrac{1}{x}\right]+1\right)=(-x)\cdot\left[-\dfrac{1}{x}\right]-x\rightarrow 1$ as $x\rightarrow 0^{-}$.

Answer 3

I’m not sure whether the interviewer will be satisfied with this:

By the substitution of $x=1/u$:

$$\lim_{x\to 0}x[1/x]=\lim_{u\to \infty}\frac{[u]}{u}$$

Since $$u-1\le[u]\le u$$ $$1-\frac{1}{u}\le\frac{[u]}{u}\le 1$$

By squeeze theorem, the required limit is $1$.

You also have to consider the case of $u \to -\infty$, but the precedure is similar.

Here is a plot of $y=[x]/x$(blue) and the green line $y=1$.

Answer 4

Correct me if wrong.

Consider $x >0$.

The unique number $n$, $n \in \mathbb{N}$,

defines a function:

$\star) $ $\left \lfloor {1/x} \right\rfloor : = n\le 1/x \lt n+1.$

$n = \left\lfloor{1/x}\right\rfloor \le 1/x \lt n+1$,or

$xn =x \left\lfloor{1/x}\right\rfloor \le x(1/x) \lt x(n+1)$.

$\star) $ implies $x >1/(n+1):$

Combining:

$[1/(n+1)]n \lt xn= x\left\lfloor{1/x}\right\rfloor \le 1.$

$\lim_{x \rightarrow 0^+} \left\lfloor{1/x}\right\rfloor =\infty$,

implies $\lim_{n \rightarrow \infty} [1/(n+1)]n =1.$

Squeeze.

The case $x \lt 0$ remains to be shown.