Question Number 34374 by rahul 19 last updated on 05/May/18
![lim_(x→1) {1−x+[x−1]+[1−x]} = ? [.]= greatest integer function.](https://www.tinkutara.com/question/Q34374.png)
$${lim}_{{x}\rightarrow\mathrm{1}} \left\{\mathrm{1}−{x}+\left[{x}−\mathrm{1}\right]+\left[\mathrm{1}−{x}\right]\right\}\:=\:? \\ $$$$\left[.\right]=\:{greatest}\:{integer}\:{function}. \\ $$
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](https://www.tinkutara.com/question/Q34377.png)
$$\left[{x}−\mathrm{1}\right]=\left[{x}\right]−\mathrm{1} \\ $$$$\left[\mathrm{1}−{x}\right]=\left[−{x}\right]+\mathrm{1} \\ $$$$\mathrm{1}−{x}+\left[{x}−\mathrm{1}\right]+\left[\mathrm{1}−{x}\right]=\mathrm{1}−{x}+\left[{x}\right]+\left[−{x}\right] \\ $$$$\left[{x}\right]+\left[−{x}\right]=\begin{cases}{\mathrm{0};\:{x}\in\mathbb{Z}}\\{−\mathrm{1};\:{x}\in\mathbb{R}\backslash\mathbb{Z}}\end{cases} \\ $$$${f}\left({x}\right)=\mathrm{1}−{x}+\left[{x}\right]+\left[−{x}\right] \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{but} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=−\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line}\:{y}=−{x} \\ $$$$\mathrm{except}\:\mathrm{for}\:{x}\in\mathbb{Z}\:\mathrm{where}\:\mathrm{it}'\mathrm{s}\:{y}=\mathrm{1}−{x} \\ $$
Commented by MJS last updated on 05/May/18
$$\mathrm{yes}.\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{from}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{is} \\ $$$$−\mathrm{1}\:\mathrm{although}\:\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{function}\:\mathrm{at}\:\mathrm{this}\:\mathrm{point}. \\ $$
Commented by rahul 19 last updated on 05/May/18
$${we}\:{are}\:{watching}\:{in}\:{the}\:{neighbourhood} \\ $$$${of}\:{x}=\mathrm{1}\:{hence}\:{ans}.\:{is}\:−\mathrm{1}. \\ $$$${right}\:? \\ $$
Commented by rahul 19 last updated on 05/May/18
$$\mathscr{T}{hank}\:{you}\:{sir}. \\ $$
Commented by rahul 19 last updated on 05/May/18
$${exactly}! \\ $$
Answered by math khazana by abdo last updated on 05/May/18
![let put x−1=t lim_(x→1) 1−x +[x−1] +[1−x] =lim_(t→0) −t +[t] +[−t] let put f(t)=−t +[t] +[−t] we have f(0)=0 but lim_(t→0^+ ) f(t) = −1 and lim_(t→0^− ) f(t)= −1 so limits exist at left and right but f is not continue at 0 .](https://www.tinkutara.com/question/Q34405.png)
$${let}\:{put}\:{x}−\mathrm{1}={t}\:\:\:{lim}_{{x}\rightarrow\mathrm{1}} \mathrm{1}−{x}\:+\left[{x}−\mathrm{1}\right]\:+\left[\mathrm{1}−{x}\right] \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:−{t}\:\:+\left[{t}\right]\:+\left[−{t}\right]\:{let}\:{put}\:{f}\left({t}\right)=−{t}\:+\left[{t}\right]\:+\left[−{t}\right] \\ $$$${we}\:{have}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:\:{but}\: \\ $$$${lim}_{{t}\rightarrow\mathrm{0}^{+} } \:\:\:{f}\left({t}\right)\:=\:−\mathrm{1}\:\:\:{and}\:{lim}_{{t}\rightarrow\mathrm{0}^{−} } \:\:{f}\left({t}\right)=\:\:−\mathrm{1}\:{so} \\ $$$${limits}\:{exist}\:{at}\:{left}\:{and}\:{right}\:\:{but}\:{f}\:{is}\:{not}\:{continue} \\ $$$${at}\:\mathrm{0}\:. \\ $$