Arturo O. answered 02/06/17
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Another way to solve this is to apply L'Hopital's rule twice to find the limit as x→0.
First application:
Limit as x→0 of
(2x + sinx + xcosx) / (2x) is still indeterminate.
Second application:
Limit as x→0 of
(2 + 2cosx - xsinx) / 2 = (2 + 2)/2 = 4/2 = 2
Limit as x→0 of
(2 + 2cosx - xsinx) / 2 = (2 + 2)/2 = 4/2 = 2