The correct option is D Does not exist because the left hand limit is not equal to the right hand limit.
L.H.L=limx→1−√1−cos[2(x−1)]x−1=limx→1−√2sin2(x−1)x−1=√2.limx→1−√sin2(x−1)x−1=√2limx→1−|sin(x−1)x−1=√2limh→0|sin(−h)|−h=√2limh→0sinh−h=−√2
Again,
R.H.L.=limx→1+√1−cos(x−1)x−1=limx→1+√2|sin(x−1)|x−1
Put x=1+h,h>0forx→1+,h→0.=limh→0√2|sinh|h=limh→0√2sinhh=√2
L.H.L ≠ R.H.L Therefore limx→1f(x) does not exist.