1.设$\int_a^{ + \infty } {f\left( x \right)dx}
$收敛,若$\mathop {\lim }\limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty }
\end{array}} f\left( x \right)$存在,则\[\mathop {\lim }\limits_{x \to
\begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} f\left( x \right) = 0\]
证明：反证法,若$\mathop {\lim }\limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty
} \end{array}} f\left( x \right) = l \ne 0$,则不妨设$l > 0$,从而 由极限的保号性知,存在$M >
0$,当$x>M$时,有\[f\left( x \right) > \frac{1}{2}l\]于是\[\mathop {\lim
}\limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}}
\int_M^x {f\left( t \right)dt} \ge \mathop {\lim }\limits_{x \to
\begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} \frac{1}{2}l\left( {x -
M} \right) = + \infty \]这与$\int_a^{ + \infty } {f\left( x \right)dx}
$收敛矛盾,故$l=0$注1：由于$\int_a^{ + \infty } {f\left( x \right)dx} $收敛,所以$\mathop {\lim
}\limits_{A \to \begin{array}{*{20}{c}} { + \infty } \end{array}} \int_a^A
{f\left( x \right)dx} $存在 注2：若$\sum\limits_{n = 1}^\infty {{a_n}} $收敛,则$\mathop
{\lim }\limits_{n \to \infty } {a_n} = 0$关于反常积分收敛与极限的专题讨论,布布扣,bubuko.com
关于反常积分收敛与极限的专题讨论
原文：http://www.cnblogs.com/ly142857/p/3664523.html
