lim_(x→1) {1−x+[x−1]+[1−x]} = ? [.]= greatest integer function.


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Question Number 34374 by rahul 19 last updated on 05/May/18
$lim_(x→1) {1−x+[x−1]+[1−x]} = ? [.]= greatest integer function.$
${l i m}_{x \to 1} {1 - x + [x - 1] + [1 - x]} = ?$ $[.] = g r e a t e s t i n t e g e r f u n c t i o n .$
	Answered by MJS last updated on 05/May/18
	$[x−1]=[x]−1 [1−x]=[−x]+1 1−x+[x−1]+[1−x]=1−x+[x]+[−x] [x]+[−x]= { ((0; x∈Z)),((−1; x∈R\Z)) :} f(x)=1−x+[x]+[−x] f(1)=0 but lim_(x→1) f(x)=−1 the graph of f(x) is the straight line y=−x except for x∈Z where it′s y=1−x$
	$[x - 1] = [x] - 1$ $[1 - x] = [- x] + 1$ $1 - x + [x - 1] + [1 - x] = 1 - x + [x] + [- x]$ $[x] + [- x] = {\begin{cases} 0; x \in Z \\ - 1; x \in R ∖ Z \end{cases}$ $f (x) = 1 - x + [x] + [- x]$ $f (1) = 0$ $but$ $\lim_{x \to 1} f (x) = - 1$ $the graph of f (x) is the straight line y = - x$ $except for x \in Z where {it}^{'} s y = 1 - x$
	Commented by MJS last updated on 05/May/18

	$yes . in this case the limit from both sides is$ $- 1 although {that}^{'} s not the value of the$ $function at this point .$
	Commented by rahul 19 last updated on 05/May/18

	$w e a r e w a t c h i n g i n t h e n e i g h b o u r h o o d$ $o f x = 1 h e n c e a n s . i s - 1 .$ $r i g h t ?$
	Commented by rahul 19 last updated on 05/May/18

	$T h a n k y o u s i r .$
	Commented by rahul 19 last updated on 05/May/18

	$e x a c t l y!$
	Answered by math khazana by abdo last updated on 05/May/18

	$l e t p u t x - 1 = t {l i m}_{x \to 1} 1 - x + [x - 1] + [1 - x]$ $= {l i m}_{t \to 0} - t + [t] + [- t] l e t p u t f (t) = - t + [t] + [- t]$ $w e h a v e f (0) = 0 b u t$ ${l i m}_{t \to 0^{+}} f (t) = - 1 a n d {l i m}_{t \to 0^{-}} f (t) = - 1 s o$ $l i m i t s e x i s t a t l e f t a n d r i g h t b u t f i s n o t c o n t i n u e$ $a t 0 .$
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