Question

用拉格朗日中值定理证明 当x>0时，ln{[(e^x)-1]/x}<x

Answer 1

Solution:
ln{[(e^x)-1]/x}<x <=> [(e^x)-1]/e^x < x
Considering function f(x)= [(e^x)-1]/e^x on the interval [0,x] , f(x) is continuous on [0,x] and differenciable on (0,x)
Hence, [f(x)-f(0)]/[x-0]=f '(c) where c is on (0,x)
Note that f '(x)=1/ e^x <1 (since x >0 )
Hence, [f(x)-f(0)]/[x-0]<1 <=> [(e^x)-1]/e^x < x <=> ln{[(e^x)-1]/x}<x when x >0.

Answer 2

设f(x)=e^x，则存在柯西属于（0，x），使得f"(柯西)=[f(x)-f(0)]/[x-0],e^(柯西)=[e^x-1]/x<e^x,两边取对数即得。