Let f (x) have a function of second order, and f '' (x) > 0, limx tends to 0f (x) / x = 1. It is proved that when x > 0, f (x) > X

By the condition, f (0) = Lim f (x) = Lim f (x) / X * Lim x = 1 * 0 = 0
And f '(0) = LIM (f (x) - f (0)) / x = Lim f (x) / x = 1
All the above limits are that x tends to 0
Because f '' (x) > 0, f '(x) is strictly increasing, so f' (x) > F '(0) = 1,
Let g (x) = f (x) - x, G '(x) = f' (x) - 1 > 0,
G (x) is increasing, so g (x) > G (0) = f (0) - 0 = 0
F (x) > x, when x > 0

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