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
Find the value of limx→ 0|x|x
If f(x)={x,when 0≤ x ≥ 1 2−x,2-x when 1 ≤ x ≥ 2 then limx→1 f(x) =