※ 引述《johnnyzsefb (AJ)》之銘言：
: II.
:
: 2.(14%) If f:R→R satisfies f(x+y)=f(x)f(y) for all x,y belong to R and
: f(x)=1+xg(x) where lim g(x)=1.
: x→0
: Show that f is differentiable,and find f(x).
: 以上
: 感謝~XD
f(0)=f(0+0)=f(0)f(0) ==> f(0)=0 or 1.
If f(0)=0, then f(x)=0 for all x.
f(x) = 1+xg(x) with g(x)→1 when x→0.
故 f(x)≠0, at least when x mear but not equal to 0.
故 f(0)≠0, 即 f(0)=1.
故
f(x)-f(0)
----------- = g(x)→1 when x→0
x
這證明了 f'(0) 存在, 且為 1.
由 f'(0) 存在及函數方程式 f(x+y)=f(x)f(y) 易證 f
處處可微, 且因此可得 f(x)=e^x.
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