15楼: Originally posted by BWM008 at 2017-08-22 20:52:22

19楼: Originally posted by ylsxz2012 at 2017-08-22 22:55:15
如果函数可导，你的证法很不错。但题目中没有说函数f在0点和其它点的导数存在，只说连续。...

12楼: Originally posted by alober at 2017-08-22 18:22:14
只要取有理序列 q_n 收敛到 x，
f(x) = \lim_{n\to\infty}f(q_n) = \lim_{n\to\infty}kq_n = k\lim_{n\to\infty}q_n = kx

15楼: Originally posted by BWM008 at 2017-08-22 20:52:22

18楼: Originally posted by peterflyer at 2017-08-22 22:18:19
令u=x+y
由f(x+y)=f(x)+f(y)，
令x=y=0,得到: f(0)=0
再对上式求导：
df(u)/du=df/dx
由于左边是x和y的函数，右边是x的函数，两者相等，必然等于一个常数，设其为k。
df(u)/du=df(x)/dx≡k
由此得到：f(x)=k ...

17楼: Originally posted by wurongjun at 2017-08-22 22:12:18
建议你看看裴礼文的--数学分析典型题解--
上面有详细的证明!