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If f (x) has second derivative in the field of x = 0, and if x → 0, LIM ((SiNx + XF (x)) \ \ x3) = 0, find f (0), f '(0), f' '(0)
The limit of 0 can only be 0 / 0 type (denominator 0) cosx + F (x) + XF '(x) / 3x2 = 0-sinx + F' (x) + F '(x) / 6x = 0-cosx + F' (x) + F '(x) + F' (x) + F '(x) / 6 = 0, all of which are carried into Formula 1 with x = 0 to get f (0) = - 12 to get f' (0) = 03 to get f '(0) = 1 / 3
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