find-lim-x-0-x-1-x-and-lim-x-0-x-2-1-x-is-the-greatest-integr-inferior-or-equal-to-

Question Number 29156 by abdo imad last updated on 04/Feb/18

f i n d {l i m}_{x \to 0^{+}} x [\frac{1}{x}] a n d {l i m}_{x \to 0^{+}} x^{2} [\frac{1}{x}] .

[α] i s t h e g r e a t e s t i n t e g r i n f e r i o r o r e q u a l t o α .

Commented by abdo imad last updated on 09/Feb/18

w e h a v e [t] ⩽ t < [t] + 1 \Rightarrow [\frac{1}{x}] ⩽ \frac{1}{x} < [\frac{1}{x}] + 1

\Rightarrow \forall x > 0 x [\frac{1}{x}] ⩽ 1 < x [\frac{1}{x}] + x \Rightarrow 0 ⩽ 1 - x [\frac{1}{x}] < x

b u t {l i m}_{x \to 0^{+}} x = 0 \Rightarrow {l i m}_{x \to 0^{+}} x [\frac{1}{x}] = 1 .

Commented by abdo imad last updated on 09/Feb/18

f o r x > 0 w e h a v e 0 ⩽ 1 - x [\frac{1}{x}] < x \Rightarrow 0 ⩽ x - x^{2} [\frac{1}{x}] < x^{2}

\Rightarrow - x ⩽ - x^{2} [\frac{1}{x}] < x^{2} - x \Rightarrow x - x^{2} < x^{2} [\frac{1}{x}] ⩽ x b u t

{l i m}_{x \to 0} (x - x^{2}) = {l i m}_{x \to 0} x = 0 \Rightarrow {l i m}_{x \to 0} x^{2} [\frac{1}{x}] = 0