(1−x)+sin(1−x)[1−x]sin(π2[1−x])=limx→1 (1−x)+sin(1−x)sin(π2[1−x])=0

" /> (1−x)+sin(1−x)[1−x]sin(π2[1−x])=limx→1 (1−x)+sin(1−x)sin(π2[1−x])=0

" />
devarshi-dt-logo

Question:

For each t∈R, let [t] be the greatest integer less than or equal to t. Then, limx→1 [(1−|x|+sin|1−x|)sin(π2[1−x])] / [(1−x)[1−x]]. Equals

Equals

Does not exist

Equals 0

Equals 1

Solution:

limx→1 [(1−|x|+sin|1−x|)sin(π2[1−x])] / [(1−x)[1−x]]=limx→1 (1−x)+sin(1−x)[1−x]sin(π2[1−x])=limx→1 (1−x)+sin(1−x)sin(π2[1−x])=0