1.5 函数极限的运算法则

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文章目录

  • 函数极限的四则运算法则
  • 复合函数的极限运算法则

函数极限的四则运算法则

  1. 函数极限的四则运算法则

定理1:如果有 lim ⁡ x → x 0 f ( x ) = A \lim_{x \rightarrow x_{0}}f(x)=A limxx0f(x)=A, lim ⁡ x → x 0 g ( x ) = B \lim_{x \rightarrow x_{0}}g(x)=B limxx0g(x)=B, 则有:
(1) lim ⁡ x → x 0 [ f ( x ) ± g ( x ) ] \lim_{x \rightarrow x_{0}}[f(x)\pm g(x)] limxx0[f(x)±g(x)] 存在,且 lim ⁡ x → x 0 [ f ( x ) ± g ( x ) ] = \lim_{x \rightarrow x_{0}}[f(x)\pm g(x)]= limxx0[f(x)±g(x)]= lim ⁡ x → x 0 f ( x ) ± \lim_{x \rightarrow x_{0}}f(x)\pm limxx0f(x)± lim ⁡ x → x 0 g ( x ) \lim_{x \rightarrow x_{0}}g(x) limxx0g(x) = A ± B A\pm B A±B;
(2) lim ⁡ x → x 0 [ f ( x ) ⋅ g ( x ) ] \lim_{x \rightarrow x_{0}}[f(x)\cdot g(x)] limxx0[f(x)g(x)] 存在,且 lim ⁡ x → x 0 [ f ( x ) ⋅ g ( x ) ] = \lim_{x \rightarrow x_{0}}[f(x)\cdot g(x)]= limxx0[f(x)g(x)]= lim ⁡ x → x 0 f ( x ) ⋅ \lim_{x \rightarrow x_{0}}f(x)\cdot limxx0f(x) lim ⁡ x → x 0 g ( x ) \lim_{x \rightarrow x_{0}}g(x) limxx0g(x) = A B AB AB;
(3) lim ⁡ x → x 0 f ( x ) g ( x ) \lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)} limxx0g(x)f(x) 存在,若 lim ⁡ x → x 0 g ( x ) = B ≠ 0 \lim_{x \rightarrow x_{0}} g(x)=B\neq 0 limxx0g(x)=B=0, 则 lim ⁡ x → x 0 f ( x ) g ( x ) = lim ⁡ x → x 0 f ( x ) lim ⁡ x → x 0 g ( x ) = A B \lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=\frac{\lim_{x \rightarrow x_{0}}f(x)}{\lim_{x \rightarrow x_{0}}g(x)}=\frac{A}{B} limxx0g(x)f(x)=limxx0g(x)limxx0f(x)=BA .

注: 使用四则运算法则的前提是各部分极限必须存在.

推论1 lim ⁡ x → x 0 [ c ⋅ f ( x ) ] = c ⋅ lim ⁡ x → x 0 f ( x ) \lim_{x \rightarrow x_{0}}[c\cdot f(x)]=c\cdot\lim_{x \rightarrow x_{0}}f(x) limxx0[cf(x)]=climxx0f(x) ( c c c 为常数).
推论2 lim ⁡ x → x 0 [ f ( x ) ] n = [ lim ⁡ x → x 0 f ( x ) ] n \lim_{x \rightarrow x_{0}}[ f(x)]^{n}=[\lim_{x \rightarrow x_{0}}f(x)]^{n} limxx0[f(x)]n=[limxx0f(x)]n ( n n n 为正整数).

  1. 数列极限的四则运算法则

定理2: 设有数列 { x n } \{x_{n}\} {xn} { y n } \{y_{n}\} {yn} , 如果 lim ⁡ n → ∞ x n = A \lim_{n \rightarrow \infty}x_{n}=A limnxn=A, lim ⁡ n → ∞ y n = B \lim_{n \rightarrow \infty}y_{n}=B limnyn=B, 则有:
(1) lim ⁡ n → ∞ ( x n ± y n ) = A ± B \lim_{n \rightarrow \infty}(x_{n}\pm y_{n})=A\pm B limn(xn±yn)=A±B;
(2) lim ⁡ n → ∞ ( x n ⋅ y n ) = A ⋅ B \lim_{n \rightarrow \infty}(x_{n}\cdot y_{n})=A\cdot B limn(xnyn)=AB;
(3)当 y n ≠ 0 ( n = 1 , 2 , . . . ) y_{n}\neq0(n=1,2,...) yn=0(n=1,2,...) B ≠ 0 B\neq0 B=0 时, lim ⁡ n → ∞ x n y n = A B \lim_{n \rightarrow \infty}\frac{x_{n} }{y_{n}}=\Large\frac{A}{B} limnynxn=BA.

  1. 例子

例1:设 n n n 次多项式 P n ( x ) = a 0 + a 1 x + a 2 x 2 + . . . + a n x n P_{n}(x)=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n} Pn(x)=a0+a1x+a2x2+...+anxn, 其中 a 0 , a 1 , a 2 , . . . , a n a_{0},a_{1}, a_{2},...,a_{n} a0,a1,a2,...,an 为多项式系数,试证: lim ⁡ x → x 0 P n ( x ) = P n ( x 0 ) \lim_{x \rightarrow x_{0}}P_{n}(x)=P_{n}(x_{0}) limxx0Pn(x)=Pn(x0).
lim ⁡ x → x 0 P n ( x ) = lim ⁡ x → x 0 ( a 0 + a 1 x + a 2 x 2 + . . . + a n x n ) \lim_{x \rightarrow x_{0}}P_{n}(x)=\lim_{x \rightarrow x_{0}} (a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}) limxx0Pn(x)=limxx0(a0+a1x+a2x2+...+anxn)
               = lim ⁡ x → x 0 a 0 + lim ⁡ x → x 0 a 1 x + lim ⁡ x → x 0 a 2 x 2 + . . . + lim ⁡ x → x 0 a n x n ~~~~~~~~~~~~~~=\lim_{x \rightarrow x_{0}}a_{0}+\lim_{x \rightarrow x_{0}}a_{1}x+\lim_{x \rightarrow x_{0}}a_{2}x^{2}+...+\lim_{x \rightarrow x_{0}}a_{n}x^{n}               =limxx0a0+limxx0a1x+limxx0a2x2+...+limxx0anxn
               = lim ⁡ x → x 0 a 0 + a 1 lim ⁡ x → x 0 x + a 2 ( lim ⁡ x → x 0 x ) 2 + . . . + a n ( lim ⁡ x → x 0 x ) n ~~~~~~~~~~~~~~=\lim_{x \rightarrow x_{0}}a_{0}+a_{1}\lim_{x \rightarrow x_{0}}x+a_{2}(\lim_{x \rightarrow x_{0}}x)^{2}+...+a_{n}(\lim_{x \rightarrow x_{0}}x)^{n}               =limxx0a0+a1limxx0x+a2(limxx0x)2+...+an(limxx0x)n
               = a 0 + a 1 x 0 + a 2 x 0 2 + . . . + a n x 0 n = P n ( x 0 ) ~~~~~~~~~~~~~~= a_{0}+a_{1}x_{0}+a_{2}x_{0}^{2}+...+a_{n}x_{0}^{n}=P_{n}(x_{0})               =a0+a1x0+a2x02+...+anx0n=Pn(x0).

例2:设 有理分式函数 R ( x ) = P ( x ) Q ( x ) R(x)=\frac{P(x)}{Q(x)} R(x)=Q(x)P(x), 其中 P ( x ) P(x) P(x) Q ( x ) Q(x) Q(x) 都是多项式,且 Q ( x 0 ) ≠ 0 Q(x_{0})\neq0 Q(x0)=0 , 试证: lim ⁡ x → x 0 R ( x ) = R ( x 0 ) \lim_{x \rightarrow x_{0}}R(x)=R(x_{0}) limxx0R(x)=R(x0).
:因为 P ( x ) P(x) P(x) Q ( x ) Q(x) Q(x) 都是多项式,由例1结论可知
lim ⁡ x → x 0 P ( x ) = P ( x 0 ) , lim ⁡ x → x 0 Q ( x ) = Q ( x 0 ) , \begin{align} \lim_{x \rightarrow x_{0}}P(x)=P(x_{0}),\lim_{x \rightarrow x_{0}}Q(x)=Q(x_{0}),\nonumber \end{align} xx0limP(x)=P(x0)xx0limQ(x)=Q(x0),
又因为 Q ( x 0 ) ≠ 0 Q(x_{0})\neq 0 Q(x0)=0, 由函数极限的四则运算法则可得
lim ⁡ x → x 0 R ( x ) = lim ⁡ x → x 0 P ( x ) Q ( x ) = lim ⁡ x → x 0 P ( x ) lim ⁡ x → x 0 Q ( x ) = P ( x 0 ) Q ( x 0 ) = R ( x 0 ) . \begin{align} \lim_{x \rightarrow x_{0}}R(x)=\lim_{x \rightarrow x_{0}}\frac{P(x)}{Q(x)}=\frac{\lim_{x \rightarrow x_{0}}P(x)}{\lim_{x \rightarrow x_{0}}Q(x)}=\frac{P(x_{0})}{Q(x_{0})}=R(x_{0}).\nonumber \end{align} xx0limR(x)=xx0limQ(x)P(x)=limxx0Q(x)limxx0P(x)=Q(x0)P(x0)=R(x0).

注: 由例1和例2可得一个结论:在求多项式函数或有理分式函数当 x → x 0 x \rightarrow x_{0} xx0 时的极限时,可直接将 x 0 x_{0} x0 替代函数中的 x x x 即可得到极限值;对于有理分式函数,前提是用 x 0 x_{0} x0 替代分母中的 x x x 后分母不能为零,如果分母为零,则不能分子、分母分别取极限,而需要做一些相应处理.

对于有理函数有如下结论:当 x → ∞ x \rightarrow \infty x, 且 a 0 ≠ 0 , b 0 ≠ 0 a_{0}\neq0, b_{0}\neq0 a0=0,b0=0, 以及 m m m n n n 为非负整数时,有 lim ⁡ x → ∞ a 0 x n + a 1 x n − 1 + . . . + a n b 0 x m + b 1 x m − 1 + . . . + b m { 0 ,     n < m , a 0 b 0 ,     n = m , ∞ ,     n > m . \begin{align} \lim_{x \rightarrow \infty}\frac{a_{0}x^{n}+a_{1}x^{n-1}+...+a_{n}}{b_{0}x^{m}+b_{1}x^{m-1}+...+b_{m}}\left\{ \begin{aligned} 0 , ~~~nm. \end{aligned}\nonumber \right. \end{align} xlimb0xm+b1xm1+...+bma0xn+a1xn1+...+an 0,   n<m,b0a0,   n=m,,   n>m.

注: 对于满足上式函数类型的极限,可直接利用此结论得出结果.


复合函数的极限运算法则

  1. 复合函数的极限运算法则

定理3:设 y = f [ g ( x ) ] y=f[g(x)] y=f[g(x)] 是由函数 u = g ( x ) u=g(x) u=g(x) 与函数 u = g ( x ) u=g(x) u=g(x) 与函数 y = f ( u ) y=f(u) y=f(u) 复合而成,若 lim ⁡ x → x 0 g ( x ) = u 0 \lim_{x \rightarrow x_{0}}g(x)=u_{0} limxx0g(x)=u0, lim ⁡ u → u 0 f ( u ) = A \lim_{u \rightarrow u_{0}}f(u)=A limuu0f(u)=A, 且在 x 0 x_{0} x0 的某去心邻域内有 g ( x ) ≠ u 0 g(x)\neq u_{0} g(x)=u0, 则 lim ⁡ x → x 0 f [ g ( x ) ] = A \lim_{x \rightarrow x_{0}}f[g(x)]=A limxx0f[g(x)]=A.


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