问题:g(x)=f(x)/xx≠0g(x)=f′(0)x=0知道f(x)有二阶连续导数f(0)=0证g可导且导函数连续g(x)=f(x)/xx≠0g(x)=f′(0)x=0知道f(x)有二阶连续导数f(0)=0证g可导且导函数连续
网友采纳 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'(...
胡宗武的回答:
网友采纳 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?
蒋渭忠的回答:
网友采纳 forx≠0g(x)=f(x)/xg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)