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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.
limx→1{1−x+[x−1]+[1−x]}=?[.]=greatestintegerfunction.
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]+11−x+[x−1]+[1−x]=1−x+[x]+[−x][x]+[−x]={0;x∈Z−1;x∈R∖Zf(x)=1−x+[x]+[−x]f(1)=0butlimx→1f(x)=−1thegraphoff(x)isthestraightliney=−xexceptforx∈Zwhereit′sy=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.
yes.inthiscasethelimitfrombothsidesis−1althoughthat′snotthevalueofthefunctionatthispoint.
Commented by rahul 19 last updated on 05/May/18
we are watching in the neighbourhood of x=1 hence ans. is −1. right ?
wearewatchingintheneighbourhoodofx=1henceans.is−1.right?
Commented by rahul 19 last updated on 05/May/18
Thank you sir.
Thankyousir.
Commented by rahul 19 last updated on 05/May/18
exactly!
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 .
letputx−1=tlimx→11−x+[x−1]+[1−x]=limt→0−t+[t]+[−t]letputf(t)=−t+[t]+[−t]wehavef(0)=0butlimt→0+f(t)=−1andlimt→0−f(t)=−1solimitsexistatleftandrightbutfisnotcontinueat0.

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