[tex]\sqrt[100]{e} > \dfrac{101}{100} \Rightarrow \sqrt[100]{e}>1+\dfrac{1}{100} \Rightarrow e^{\dfrac{1}{100}}- \dfrac{1}{100} - 1 > 0~~ (A)? \\ \\ \text{Consideram functia }f:\mathbb{R}\to \mathbb{R},~~f(x) = e^x-x-1 \\ \\ \text{Trebuie sa demonstram ca }f\Big(\dfrac{1}{100}\Big) >0 \\ \\ f'(x) = e^x - 1 \\ f'(x) = 0 \Rightarrow e^x-1 = 0 \Rightarrow e^x = 1 \Rightarrow x = 0[/tex]
[tex]\lim\limits_{x\to -\infty}f(x) = +\infty \\ \lim\limits_{x\to +\infty} f(x) = +\infty[/tex]
[tex]\Rightarrow f(0) = f_{min} \Rightarrow f_{min} = e^0-0-1 = 1-1 = 0 \\ \\ \Rightarrow f(x) > 0,\quad \forall x\in \mathbb{R}^* \\ \\ \Rightarrow f\Big(\dfrac{1}{100}\Big) > 0 \Rightarrow \sqrt[100]{e} >\dfrac{101}{100}[/tex]
