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Question Number 105036 by bobhans last updated on 25/Jul/20

\lim_{x \to 0} x . [\frac{1}{x}] ?

n o t e [] = g r e a t e s t i n t e g e r f u n c t i o n

Answered by john santu last updated on 25/Jul/20

t h e l i m i t a s x \to 0^{+} c a n b e

t r a n s f o r m e d i n t o \lim_{p \to \infty} \frac{⌊ p ⌋}{p}

s e t {p} = p - ⌊ p ⌋

w e h a v e \frac{⌊ p ⌋}{p} = 1 - \frac{{p}}{p}, s i n c e

0 ⩽ {p} < 1 . s o \lim_{p \to \infty} (1 - \frac{{p}}{p}) =

1 - 0 = 1 . s i m i l a r l y f o r

t h e l i m i t a s x \to 0^{-}

(J S ♠ ⧫)

Answered by mathmax by abdo last updated on 25/Jul/20

we have [\frac{1}{x}] ⩽ \frac{1}{x} < [\frac{1}{x}] + 1 \Rightarrow for x > 0 \Rightarrow x [\frac{1}{x}] ⩽ 1 < x [\frac{1}{x}] + x \Rightarrow

{\begin{cases} x [\frac{1}{x}] ⩽ 1 \Rightarrow 1 - x < x [\frac{1}{x}] ⩽ 1 we passe to limit (x \to o^{+}) \Rightarrow \\ 1 - x < x [\frac{1}{x}] \end{cases}

\lim_{x \to 0^{+}} x [\frac{1}{x}] = 1