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lim x ⌊1 / x⌋ , x → 0

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I've seen it at Yahoo! Answers: find $\lim_{x\to 0} x \left\lfloor\dfrac{1}{x}\right\rfloor$.

According to the definition of the floor function, $\forall t\in\mathbb{R}$ we verify $0\le t-\lfloor t\rfloor <1$. So, if $x\neq 0$, $$0 ≤ \frac{1}{x}− \left\lfloor\frac{1}{x}\right\rfloor< 1\Rightarrow -\frac{1}{x}\le -\left\lfloor\frac{1}{x}\right\rfloor < 1-\frac{1}{x}$$ Multiplyng both sides by $x > 0:\quad$ $-1\le -x\left\lfloor\dfrac{1}{x}\right\rfloor <x-1$

Multiplying by $-1:\quad$ $1 − x < x \left\lfloor\dfrac{1}{x}\right\rfloor ≤ 1$

Applying limits: $\quad 1=\lim_{x\to 0^+} ( 1− x )\le \lim_{x\to 0^+} x \left\lfloor\dfrac{1}{x}\right\rfloor ≤ \lim_{x\to 0^+} 1=1$

This implies: $\quad \lim_{x\to 0^+} x \left\lfloor\dfrac{1}{x}\right\rfloor = 1$

With similar arguments: $\quad\lim_{x\to 0^-} x \left\lfloor\dfrac{1}{x}\right\rfloor = 1$

We can conclude that: $\quad\lim_{x\to 0} x \left\lfloor\dfrac{1}{x}\right\rfloor = 1$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Alternatively.

Substitute \(\displaystyle x=\frac 1 t\):
$$\lim_{x \to 0+} x \left\lfloor {\frac 1 x} \right\rfloor = \lim_{t \to \infty} \frac 1 t \lfloor t \rfloor$$

Since \(\displaystyle 1 = \frac 1 t (t) \le \frac 1 t \lfloor t \rfloor < \frac 1 t (t + 1) \to 1\), we get due to the squeeze theorem:
$$\lim_{x \to 0+} x \left\lfloor {\frac 1 x} \right\rfloor = 1$$