Given that f (x) is differentiable at x0, then when h approaches 0, f (x0 + H) − f (x0 − h) 2H approaches () A. 12f′(x0)B. f′（x0）C. 2f′（x0）D. 4f′（x0）

Given that f (x) is differentiable at x0, then when h approaches 0, f (x0 + H) − f (x0 − h) 2H approaches () A. 12f′(x0)B. f′（x0）C. 2f′（x0）D. 4f′（x0）

From the topic meaning, f (x0 + H) − f (x0 − h) 2H = 12 [f (x0 + H) − f (x0) H + F (x0) − f (x0 − h) H] ∵ f (x) can be derived at x0, when h tends to 0, f (x0 + H) − f (x0 − h) 2H tends to 12 [f '(x0) + F' (x0)], that is, when h tends to 0, f (x0 + H) − f (x0 − h) 2H tends to f '(x0), so B is selected

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