已知函数f（x）在R+上可导，f（x）＞0，limx→+∞f(x)=1，且满足limh→0(f(x+hx)f(x))1h=e1x，求f（x）．

Question

已知函数f（x）在R⁺上可导，f（x）＞0，

lim

x→+∞

f(x)=1，且满足

lim

h→0

((f(x+hx))/(f(x)))(1)/(h)=e(1)/(x)，求f（x）．

Answer 1

设y=((f(x+hx))/(f(x)))(1)/(h)
两边取对数得
lny=(1)/(h)ln(f(x+hx))/(f(x))
因为

lim

h→0

lny=

lim

h→0

(1)/(h)ln(f(x+hx))/(f(x))
=

lim

h→0

(x[lnf(x+hx)-lnf(x)])/(hx)
=x[lnf（x）]'
故

lim

h→0

((f(x+hx))/(f(x)))(1)/(h)=ex[lnf(x)]′
由已知条件知ex[lnf(x)]′=e(1)/(x)
因此x[lnf(x)]′=(1)/(x)
即[lnf(x)]′=(1)/(x2)
解得
f(x)=Ce-(1)/(x)
又

lim

x→+∞

f(x)=1得C=1
故 f(x)=e-(1)/(x)