How to find limit of function: $\lim_{x\to 0}\left(x{{\left\lfloor{ \frac{1}{x}} \right\rfloor}}\right)$

Question

Find $\displaystyle \lim_{x\to 0}\left(x{{\left\lfloor{ \frac{1}{x}} \right\rfloor}}\right)$.

When $x\to0^{+}$ I have $\left(x{\left\lfloor{ \frac{1}{x}} \right\rfloor}\right)\to 0\cdot \infty$.

When $\to 0^{-}$ I have $\left(x{\left\lfloor{ \frac{1}{x}} \right\rfloor}\right)\to 0\cdot \infty$.

I have no idea,

Answer 1

Letting $x\to0+$ is the same as letting ${1\over x}=:y\to\infty$, i.e., $$\lim_{x\to0+}\bigl(x\>\lfloor 1/x\rfloor\bigr)=\lim_{y\to\infty}{\lfloor y\rfloor\over y}\ .$$ Since $\ y-1<\lfloor y\rfloor\leq y$ it follows that the limit is $1$, by the squeeze theorem. The case $x\to0-$ leads to $y\to-\infty$, and is similar.

Answer 2

Hint: For $x\gt0$, $$ 1-x=x\left(\frac1x-1\right)\le x\left\lfloor\frac1x\right\rfloor\le x\left(\frac1x\right)=1 $$ The inequalities are reversed for $x\lt0$, but you can still apply the Squeeze Theorem in both cases.

Answer 3

HINT: Use the definition of the floor function. If $\frac1{n+1}<x\le\frac1n$ for some positive integer $n$, then $n\le\frac1x<n+1$, so $\left\lfloor\frac1x\right\rfloor=n$, and

$$\frac{n}{n+1}<x\left\lfloor\frac1x\right\rfloor<\frac{n+1}n\;;$$

what are the limits of $\frac{n}{n+1}$ and $\frac{n+1}n$ as $n\to\infty$?

Now do something similar for $x<0$.