Given that f (x) is continuous in a neighborhood of x = 0, and f (0) = 0, limx → 0f (x) 1-cosx = 2, then f (x) () A. Nondifferentiable B. differentiable, and f ′ (0) ≠ 0C. Get the maximum D. get the minimum

Limx → 0f (x) 1-cosx = limx → 0f (x) 12x2 + O (x2) = 2, so in the x = 0 critical field, f (x) = x2 + O (x2) f '(x) = 2x + O (x) f' '(x) = 2 + O (1), so at the point x = 0, f' (0) = 0, f '' (0) = 2 ＞ 0, that is, at the point x = 0, the function f (x) gets the minimum

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