g(x)=f(x)/x ; x≠0 =f′(0) ; x=0g'(x) = lim(y->0) [g(x+y) - g(x)] / yg'(0) = lim(y->0) [ g(y) - g(0) ]/y= lim(y->0) [ f(y)/y - f'(0) ]/y= lim(y->0) [ f(y) - yf'(0) ]/y^2 (0/0)= lim(y->0) [ f'(y) - f'(...逻辑就是说我假设g（x）可导，然后证明了导函数在0处连续，所以也可以说明它可导了，是这样么？for x≠0 g(x) = f(x)/x g 在 x≠0g'(x) 存在，那是不用质疑 g'(x) = lim(y->0) [g(x+y) - g(x)] /y(这是导数的定义) 我已经证明了它存在，而且g'(0) =f''(0)/2 = lim(x->0) g'(x)