关于极限证明方法的专题讨论I

定义法

$\bf命题：$设$\left\{ {{a_n}} \right\}$为单调增加的数列,则$\lim \limits_{n \to \infty } {a_n} = \mathop {Sup}\limits_{k \ge 1} \left\{ {{a_k}} \right\}$

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$\bf命题：$设$f(x)$在$U_ - ^ \circ \left( {{x_0}} \right)$上单调递增，则$f\left( {{x_0} - 0} \right) = \mathop {Sup}\limits_{x \in U_ - ^ \circ \left( {{x_0}} \right)} f\left( x \right)$

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$\bf命题：$设$f\left( x \right) \in {C^2}\left( {0, + \infty } \right)$，且$f\left( x \right) > 0,f'\left( x \right) \le 0,f''\left( x \right)$有界，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f'\left( x \right) = 0$

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夹逼原理

$\bf命题：$任何实数都是某个有理数列的极限

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$\bf命题：$设$f\left( x \right)$在$\left( {0,1} \right)$上单调,且无界广义积分$\int_0^1 {f\left( x \right)dx} $收敛,则\[\mathop {\lim }\limits_{n \to \infty } \frac{{f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + \cdots + f\left( {\frac{{n - 1}}{n}} \right)}}{n} = \int_0^1 {f\left( x \right)dx} \]

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$\bf命题：$设$f\left( x \right)$在$\left[ {0, + \infty } \right)$单调，且$\int_0^{ + \infty } {f\left( x \right)dx} $，则

\[\mathop {\lim }\limits_{h \to \begin{array}{*{20}{c}}
{{0^ + }}
\end{array}} h\sum\limits_{n = 1}^\infty {f\left( {nh} \right)} = \int_0^{ + \infty } {f\left( x \right)dx} \]

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$\bf命题：$设$f'\left( x \right) \in R\left[ {a,b} \right]$，令\[{A_n} = \sum\limits_{i = 1}^n {f\left( {a + \frac{{i\left( {b - a} \right)}}{n}} \right)}  \cdot \frac{{\left( {b - a} \right)}}{n} - \int_a^b {f\left( x \right)dx} \]证明：$\lim \limits_{n \to \infty } n{A_n} = \frac{{b - a}}{2}\left( {f\left( b \right) - f\left( a \right)} \right)$

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$\bf命题：$设$f(x)$为$[0,+\infty)$上单调函数，且$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty } \end{array}} f\left( x \right) = 0$，证明：\[\mathop {\lim }\limits_{n \to \infty } \int_0^{ + \infty } {f\left( x \right)\sin nxdx}  = 0\]

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$\bf命题：$

上下极限法

$\bf命题：$设正数序列${x_n}$满足${x_{n + 2}} \leqslant \frac{{{x_{n + 1}} + {x_n}}}{2}\left( {n = 1,2, \cdots } \right)$，证明：$\lim \limits_{n \to \infty } {x_n}$存在

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$\bf命题：$设正数列$\left\{ {{a_n}} \right\}$满足条件${a_{n + m}} \le {a_n}{a_m},\forall n,m \in {N_ + }$，则有\[\mathop {\lim }\limits_{n \to \infty } \frac{{\ln {a_n}}}{n} = \mathop {\inf }\limits_{n \ge 1} \left\{ {\frac{{\ln {a_n}}}{n}} \right\}\]

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$\bf命题：$设正数列$\left\{ {{a_n}} \right\}$满足条件${a_{n + m}} \le {a_n}{a_m},\forall n,m \in {N_ + }$，则极限$\lim \limits_{n \to \infty }\sqrt[n]{{{a_n}}}$一定存在

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$\bf命题：$设数列$\left\{ {{a_n}} \right\}$满足条件${a_{n + m}} \le {a_n}{\rm{ + }}{a_m},\forall n,m \in {N_ + }$，则有\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{n}{\rm{ = }}\mathop {\inf }\limits_{n \ge 1} \left\{ {\frac{{{a_n}}}{n}} \right\}\]

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$\bf命题：$设${a_n}$满足${a_m} + {a_n} - 1 \leqslant {a_{m + n}} \leqslant {a_m} + {a_n} + 1\left( {\forall m,n \geqslant 1} \right)$，证明：$\lim \limits_{n \to \infty } \frac{{{a_n}}}{n}$存在

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$\bf命题：$设正值函数$f\left( x \right) \in C\left[ {a,b} \right]$，则\[\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\int_a^b {{f^n}\left( x \right)} dx}} = \mathop {max}\limits_{x \in \left[ {a,b} \right]} f\left( x \right)\]

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$\bf命题：$设正值函数$f\left( x \right),g\left( x \right) \in C\left[ {a,b} \right]$，则\[\mathop {\lim }\limits_{n \to \infty } \frac{{\int_a^b {{f^{n + 1}}\left( x \right)g\left( x \right)dx} }}{{\int_a^b {{f^n}\left( x \right)g\left( x \right)dx} }} = \mathop {max}\limits_{x \in \left[ {a,b} \right]} f\left( x \right)\]

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$\bf命题：$