若f′(x0)f^\prime(x_0)f′(x0)存在，则limh→0f(x0+h2)−f(x0+2h)h2=\lim\limits_{h\to0}\dfrac{f(x_0 + h^{2}) - f(x_0 + 2h)}{h^{2}} =h→0limh2f(x0+h2)−f(x0+2h)=（ ）.
（A）f′(x0)−2f′(x0)f'(x_0)-2f'(x_0)f′(x0)−2f′(x0)
（B）2f′(x0)2f'(x_0)2f′(x0)
（C）−2f′(x0)-2f'(x_0)−2f′(x0)
（D）−f′(x0)-f'(x_0)−f′(x0)
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