_ x 0 [ |2 + x| - |2 - x| x ] = .......

Step-by-step solution

As x 0 , 2+x > 0 and 2-x > 0 , so the absolute value signs can be removed: _ x 0 (2 + x) - (2 - x) x This limit is of the form 0 0 . Applying L'Hospital's rule by differentiating the numerator and the denominator with respect to x : _ x 0 1 2 + x - 1 2 - x (-1) ^2 x = _ x 0 1 2 + x + 1 2 - x ^2 x Substituting x = 0 : = 1 2 + 1 2 1 = 1 Answer: 1