2018届成都市二诊数学(文)21题第二问的另类证法(切线放缩法)

题(2)当\(x\in (1,+\infty)\)时，求证：\(\dfrac{\text{e}(x-1)}{\text{e}^x}<\ln x<x^2-x\).

\(\ding{192}\)函数\(y=\dfrac{\ln x}{x}\)在\(x=1\)处切线放缩\(\Rightarrow \dfrac{\ln x}{x}\leqslant x-1\)

\(\Rightarrow \ln x<x^2-x\)(其中\(x>1\))

\(\ding{193}\)函数\(y=x \ln x\)在\(x=1\)处切线放缩\(\Rightarrow x\ln x\geqslant x-1\)

\(\Rightarrow \ln x>\dfrac{x-1}{x}>\dfrac{\text{e}(x-1)}{\text{e}^x}\)(其中\(x>1\))

其中函数\(y=\text{e}^x\)在\(x=1\)处切线放缩\(\Rightarrow \text{e}^x\geqslant\text{e}x\Rightarrow\) 当\(x>1\)时\(\text{e}^x>\text{e}x\)

其中左端还可以直接使用：函数\(y=\dfrac{\text{e}^x \ln x}{\text{e}}\)在\(x=1\)处切线放缩$ \Rightarrow \dfrac{\text{e}^x \ln x}{\text{e}} \geqslant x-1(x>1) $