limx→0 √1−cos2x√2x is
(JEE 2002)
1
-1
Zero
Does not exist
limx→0√1−cos2x√2x=limx→0√1−(1−2sin2x)√2x; limx→0√2sin2x√2x=limx→0|sin x|x The limit of above does not exist as LHS = -1 ≠ RHL = 1
limx→0(e1/x−1)(e1/x)+1
If f(x)=xsin(1x),x≠0, then limx→0f(x)=