📝 题目
例 8 求证:
(1) ${\mathrm{e}}^{x} > 1 + x\left( {x \neq 0}\right)$ ;
(2)序列 ${x}_{n} = \left( {1 + \frac{1}{2}}\right) \left( {1 + \frac{1}{{2}^{2}}}\right) \cdots \left( {1 + \frac{1}{{2}^{n}}}\right)$ 的极限存在.
💡 答案与解析
证 (1) 令 $f\left( x\right) = {\mathrm{e}}^{x} - 1 - x$ ,则有 $f\left( 0\right) = 0$ ,且
$$ {f}^{\prime }\left( x\right) = {\mathrm{e}}^{x} - 1 \Rightarrow x{f}^{\prime }\left( x\right) = x\left( {{\mathrm{e}}^{x} - 1}\right) > 0\;\left( {\forall x \neq 0}\right) $$
$$ \Rightarrow x = 0\text{ 是最小值点 } \Rightarrow f\left( x\right) > f\left( 0\right) = 0 $$
$$ \Rightarrow {\mathrm{e}}^{x} > 1 + x\;\left( {\forall x \neq 0}\right) . $$
(2)显然序列 $\left\{ {x}_{n}\right\}$ 单调递增,为了证明极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}}$ 存在,只要肯定序列 $\left\{ {x}_{n}\right\}$ 有上界即可. 为此利用第 (1) 小题,有
$$ {x}_{n} < {\mathrm{e}}^{\frac{1}{2}} \cdot {\mathrm{e}}^{\frac{1}{{2}^{2}}} \cdot {\mathrm{e}}^{\frac{1}{{2}^{3}}} \cdot \cdots \cdot {\mathrm{e}}^{{2}^{n}} = {\mathrm{e}}^{\frac{1}{2} + \frac{1}{{2}^{2}} + \frac{1}{{2}^{3}} + \cdots + \frac{1}{{2}^{n}}} < {\mathrm{e}}^{1} = \mathrm{e}. $$
\subsubsection{三、函数最值应用题}