Iff(x)=x−ex+cos2xx2,x≠0 , is continuous at x=0 , then

Ltx→0x−ex+1−(1−cos2x)x2=Ltx→0[x−ex+1x2−(1−cos2x)x2] =Ltx→0[x+1−(1+x+x22)x2−2sin2xx2]       (Using expansion of ex )=−12−2 =−52 ; hence for continuity f(0)=−52 Now [f(0)] = -3; {f(0)} =[−52]=12 Hence, [f(0)]{f(0)}=−32=−1.5