📝 题目
例 7 我们把
$$ {M}_{t}\left( x\right) = {\left( \frac{{x}_{1}^{t} + {x}_{2}^{t} + \cdots + {x}_{n}^{t}}{n}\right) }^{\frac{1}{t}} $$
称为 $n$ 个正数 ${x}_{1},{x}_{2},\cdots ,{x}_{n}$ 的 $t$ 次方平均数,并记
$$ G\left( x\right) = \sqrt[n]{{x}_{1}{x}_{2}\cdots {x}_{n}}, $$
$$ M = \max \left\{ {{x}_{1},{x}_{2},\cdots ,{x}_{n}}\right\} , $$
$$ m = \min \left\{ {{x}_{1},{x}_{2},\cdots ,{x}_{n}}\right\} . $$
试证
(1) $\mathop{\lim }\limits_{{t \rightarrow 0}}{M}_{t}\left( x\right) = G\left( x\right)$ ;
(2) $\mathop{\lim }\limits_{{t \rightarrow + \infty }}{M}_{t}\left( x\right) = M$ ;
(3) $\mathop{\lim }\limits_{{t \rightarrow - \infty }}{M}_{t}\left( x\right) = m$ .
💡 答案与解析
证明(1)中的极限是 $\displaystyle{1}^{\infty }}$ 型的. 通过取对数就可以把它转化成 $\frac{0}{0}$ 型:
$$ \ln {M}_{t}\left( x\right) = \frac{\ln \left( {{x}_{1}^{t} + {x}_{2}^{t} + \cdots + {x}_{n}^{t}}\right) - \ln n}{t}. $$
利用洛必达法则, 我们求得
$$ \mathop{\lim }\limits_{{t \rightarrow 0}}\ln {M}_{t}\left( x\right) = \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{\ln \left( {{x}_{1}^{t} + \cdots + {x}_{n}^{t}}\right) - \ln n}{t} $$
$$ = \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{{x}_{1}^{t}\ln {x}_{1} + \cdots + {x}_{n}^{t}\ln {x}_{n}}{{x}_{1}^{t} + \cdots + {x}_{n}^{t}} $$
$$ = \frac{\ln \left( {{x}_{1}{x}_{2}\cdots {x}_{n}}\right) }{n} = \ln G\left( x\right) , $$
$$ \mathop{\lim }\limits_{{t \rightarrow 0}}{M}_{t}\left( x\right) = \mathop{\lim }\limits_{{t \rightarrow 0}}{\mathrm{e}}^{\ln {M}_{t}\left( x\right) } = {\mathrm{e}}^{\ln G\left( x\right) } = G\left( x\right) ; $$
(2)不妨设
$$ {x}_{1} \leq \cdots \leq {x}_{k} < {x}_{k + 1} = \cdots = {x}_{n} = M, $$
于是
$$ \mathop{\lim }\limits_{{t \rightarrow + \infty }}\ln {M}_{t}\left( x\right) = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{\ln \left( {{x}_{1}^{t} + \cdots + {x}_{n}^{t}}\right) - \ln n}{t} $$
$$ = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{{x}_{1}^{t}\ln {x}_{1} + \cdots + {x}_{n}^{t}\ln {x}_{n}}{{x}_{1}^{t} + \cdots + {x}_{n}^{t}} $$
$$ = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{{\left( \frac{{x}_{1}}{{x}_{n}}\right) }^{t}\ln {x}_{1} + \cdots + {\left( \frac{{x}_{n}}{{x}_{n}}\right) }^{t}\ln {x}_{n}}{{\left( \frac{{x}_{1}}{{x}_{n}}\right) }^{t} + \cdots + {\left( \frac{{x}_{n}}{{x}_{n}}\right) }^{t}} $$
$$ = \frac{\left( {n - k}\right) \ln M}{n - k} = \ln M, $$
$$ \mathop{\lim }\limits_{{t \rightarrow + \infty }}{M}_{t}\left( x\right) = M $$
(3)不妨设
$$ m = {x}_{1} = \cdots = {x}_{h} < {x}_{h + 1} \leq \cdots \leq {x}_{n}, $$
于是
$$ \mathop{\lim }\limits_{{t \rightarrow - \infty }}\ln {M}_{t}\left( x\right) = \mathop{\lim }\limits_{{t \rightarrow - \infty }}\frac{\ln \left( {{x}_{1}^{t} + \cdots + {x}_{n}^{t}}\right) - \ln n}{t} $$
$$ = \mathop{\lim }\limits_{{t \rightarrow - \infty }}\frac{{x}_{1}^{t}\ln {x}_{1} + \cdots + {x}_{n}^{t}\ln {x}_{n}}{{x}_{1}^{t} + \cdots + {x}_{n}^{t}} $$
$$ = \mathop{\lim }\limits_{{t \rightarrow - \infty }}\frac{{\left( \frac{{x}_{1}}{{x}_{1}}\right) }^{t}\ln {x}_{1} + \cdots + {\left( \frac{{x}_{n}}{{x}_{1}}\right) }^{t}\ln {x}_{n}}{{\left( \frac{{x}_{1}}{{x}_{1}}\right) }^{t} + \cdots + {\left( \frac{{x}_{n}}{{x}_{1}}\right) }^{t}} $$
$$ = \frac{h\ln m}{h} = \ln m, $$
$$ \mathop{\lim }\limits_{{t \rightarrow - \infty }}{M}_{t}\left( x\right) = m $$