【高等数学笔记-极限(4)】极限的运算法则
运算法则
有关无穷小量的:
- 两个无穷小量之和为无穷小量
- 两个无穷小量之积为无穷小量
- 有限个无穷小量之和为无穷小量
- 有限个无穷小量之积为无穷小量
- 常数乘以无穷小量为无穷小量
- 有界函数与无穷小量之积为无穷小量
不可使用的:
- 两个无穷小量之商:不定式
注意必须2个极限都存在时才可以使用,否则不可以使用!!并且这里的极限代表不同极限情况(趋于某点或者趋于无穷)
设limf(x)=A,limg(x)=B\lim f(x)=A,\lim g(x)=Blimf(x)=A,limg(x)=B,那么有:
- lim[f(x)+g(x)]=limf(x)+limg(x)\lim [f(x)+g(x)]=\lim f(x)+\lim g(x)lim[f(x)+g(x)]=limf(x)+limg(x)
- lim[f(x)−g(x)]=limf(x)−limg(x)\lim [f(x)-g(x)]=\lim f(x)-\lim g(x)lim[f(x)−g(x)]=limf(x)−limg(x)
- lim[f(x)⋅g(x)]=limf(x)⋅limg(x)\lim [f(x)\cdot g(x)]=\lim f(x)\cdot \lim g(x)lim[f(x)⋅g(x)]=limf(x)⋅limg(x)
- limf(x)g(x)=limf(x)limg(x),g(x)≠0\lim \frac{f(x)}{g(x)}=\frac{\lim f(x)}{\lim g(x)} ,\quad g(x)\ne 0limg(x)f(x)=limg(x)limf(x),g(x)=0
这个四则运算同样适用于数列的极限运算
即同一极限过程中,函数之和(差,积,商)的极限,等于函数的极限之和(差,积,商);
证明:
即证明当limf(x)=A,limg(x)=B\lim f(x)=A,\lim g(x)=Blimf(x)=A,limg(x)=B时,
有lim[f(x)+g(x)]=limf(x)+limg(x)=A+B\lim [f(x)+g(x)]=\lim f(x)+\lim g(x)=A+Blim[f(x)+g(x)]=limf(x)+limg(x)=A+B
即证明f(x)+g(x)f(x)+g(x)f(x)+g(x)的极限是A+BA+BA+B
∀ε>0,∃δ>0;当0<∣x−x0∣<δ时,有∣f(x)+g(x)−(A+B)∣<ε\forall \varepsilon>0,\exists \delta>0;当0<|x-x_0|<\delta时,有|f(x)+g(x)-(A+B)|<\varepsilon∀ε>0,∃δ>0;当0<∣x−x0∣<δ时,有∣f(x)+g(x)−(A+B)∣<ε
有
A=limf(x)=f(x)+α;B=limf(x)=f(x)+β A=\lim f(x)=f(x)+\alpha;B=\lim f(x)=f(x)+\beta A=limf(x)=f(x)+α;B=limf(x)=f(x)+β
∣f(x)+g(x)−(A+B)∣=∣A−α+B−β−(A+B)∣ ⟹ ∣α+β∣<ε |f(x)+g(x)-(A+B)|=|A-\alpha+B-\beta - (A+B)| \implies |\alpha+\beta |<\varepsilon ∣f(x)+g(x)−(A+B)∣=∣A−α+B−β−(A+B)∣⟹∣α+β∣<ε
得证
- lim[c⋅f(x)]=c⋅limf(x)\lim [c\cdot f(x)]=c\cdot \lim f(x)lim[c⋅f(x)]=c⋅limf(x)
注意求极限的变量对象是谁,非自变量都视为常数ccc
例如
limx→2x2n+1=1(n+1)⋅limx→2x2=4(n+1)\lim_{x\to 2} \frac{x^2}{n+1}=\frac{1}{(n+1)}\cdot \lim_{x\to 2} {x^2}=\frac{4}{(n+1)}x→2limn+1x2=(n+1)1⋅x→2limx2=(n+1)4
limn→2x2n+1=x2⋅limn→21(n+1)=x23\lim_{n\to 2} \frac{x^2}{n+1}=x^2\cdot \lim_{n\to 2} {\frac{1}{(n+1)}}=\frac{x^2}{3}n→2limn+1x2=x2⋅n→2lim(n+1)1=3x2
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lim[f(x)]n=[limf(x)]n,n∈N+\lim [f(x)]^n=[\lim f(x)]^n,n \in \mathbb{N^+}lim[f(x)]n=[limf(x)]n,n∈N+
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φ(x)≥ψ(x) ⟹ limφ(x)≥ψ(x)\varphi(x)\ge\psi(x)\implies \lim{\varphi(x)}\ge{\psi(x)}φ(x)≥ψ(x)⟹limφ(x)≥ψ(x)
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多项式的分数形式的商(有理函数),x→∞x\to\inftyx→∞时极限,最高次数的系数之比就是极限;如果分母最高次次数比分子高,则极限无穷大;分母的最高次次数大于分子,则极限无穷小;
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复合函数的极限法则,即先求内层的极限,再求外层极限;举例:
limx→0esinx ⟹ limx→0sinx⏟u=0 ⟹ limu→0eu=1 \lim_{x \to 0}e^{\sin x} \implies \lim_{x \to 0}{\underbrace{\sin x}_{u}}=\boxed{0}\implies \lim_{u \to \boxed{0}}{e^u}=1 x→0limesinx⟹x→0limusinx=0⟹u→0limeu=1
例题
有理函数极限
趋于某点的极限x→x0x\to x_0x→x0
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最简单情况:代入x0x_0x0后,分母不为0,分子不为0,直接代入求值
limx→2x2+5x−3=−9 \lim_{x\to 2}{\frac{x^2+5}{x-3}}=-9 x→2limx−3x2+5=−9 -
代入x0x_0x0后,分母不为0,分子为0;实际就是0
limx→2x−2x+2=04=0 \lim_{x\to 2}{\frac{x-2}{x+2}}=\frac{0}{4}=0 x→2limx+2x−2=40=0 -
代入x0x_0x0后,分母为0,分子是非零常数,即1/x1/x1/x形式;
limx→2x3+2x2(x−2)2=∞ \lim_{x\to 2}{\frac{x^3+2x^2}{(x-2)^2}}=\infty x→2lim(x−2)2x3+2x2=∞ -
代入x0x_0x0后,分母,分子均为0,需要具体分析
limx→4x2−6x+8x2−5x+4=limx→4(x−2)(x−4)(x−4)(x−1)=limx→4x−2x−1=23 \begin{align*} &\lim_{x\to 4}{\frac{x^2-6x+8}{x^2-5x+4}}\\ =&\lim_{x\to 4}{\frac{(x-2)(x-4)}{(x-4)(x-1)}}\\ =&\lim_{x\to 4}{\frac{x-2}{x-1}}\\ =&\frac{2}{3} \end{align*} ===x→4limx2−5x+4x2−6x+8x→4lim(x−4)(x−1)(x−2)(x−4)x→4limx−1x−232
趋于x→∞x\to \inftyx→∞的极限
- 取最高次系数比,就是极限;如果分子分母不同次,就是∞\infty∞或者000
limx→∞x22x+1=∞ \lim_{x\to \infty}{\frac{x^2}{2x+1}}=\infty x→∞lim2x+1x2=∞
limx→∞2x+1x2=0 \lim_{x\to \infty}{\frac{2x+1}{x^2}}=0 x→∞limx22x+1=0
limx→∞2x2+2x+13x2+10x+5=23 \lim_{x\to \infty}{\frac{2x^2+2x+1}{3x^2+10x+5}}=\frac{2}{3} x→∞lim3x2+10x+52x2+2x+1=32
总结一下,当出现了求分数形式多项式极限(有理函数极限);
- x→x0x\to x_0x→x0
- 先直接代入x0x_0x0:若分母不为0,代入x0x_0x0求函数值就是极限值;
- 直接代入x0x_0x0值,若得到分母为0
- 若分子不为0,则极限为∞\infty∞,符号需要具体分析
- 尝试化简式子,看看化简后直接代入分母是否还为0,不为0了就可直接代入;化简的核心目的,就是将使得分母出现0的表达式提取出来,看是否可以与分子约分;
- x→∞x\to \inftyx→∞
- 最高次数的系数之比就是极限;如果分母最高次次数比分子高,则极限无穷大;分母的最高次次数大于分子,则极限是0;
求幂指函数(底数和指数均含有变量xxx)的极限
limn→∞(nn+1)2n=limn→∞nn+1limn→∞2n=10=1 \begin{align*} &\lim_{n\to \infty}{\left(\frac{n}{n+1}\right)^{\frac{2}{n}}}\\ =&\lim_{n\to \infty}{\frac{n}{n+1}^{\lim_{n\to \infty}{\frac{2}{n}}}}\\ =&1^0=1\\ \end{align*} ==n→∞lim(n+1n)n2n→∞limn+1nlimn→∞n210=1
limx→∞(2x2−1x2+1)x2=limx→∞(2x2−1x2+1)limx→∞x2=limu→+∞2u=+∞ \begin{align*} &\lim_{x\to \infty}{\left(\frac{2x^2-1}{x^2+1}\right)^{x^2}}\\ =&\lim_{x\to \infty}{\left(\frac{2x^2-1}{x^2+1}\right)}^{\lim_{x\to \infty}x^2}\\ =&\lim_{u\to+\infty}2^{u}=+\infty \end{align*} ==x→∞lim(x2+12x2−1)x2x→∞lim(x2+12x2−1)limx→∞x2u→+∞lim2u=+∞
- 分别求底数和指数的极限;
- 若底数极限是大于0的常数A,A≠1A,A\ne1A,A=1:
- 指数极限是非000常数BBB,则极限为ABA^BAB
- 指数极限为000,极限为1
- 指数极限为∞\infty∞:
- 0<底数极限<1:
- 指数极限为+∞+\infty+∞,极限为0
- 指数极限为−∞-\infty−∞,极限为+∞+\infty+∞
- 底数极限>1:
- 指数极限为−∞-\infty−∞,极限为0
- 指数极限为+∞+\infty+∞,极限为+∞+\infty+∞
- 以下特殊情况,需要单独讨论
- 底数极限为1时
- 指数极限为不确定方向的∞\infty∞,例如limx→01x\lim_{x\to 0}{\frac{1}{x}}limx→0x1
等差,等比公式解决
limn→∞(1n3+2n3+⋯+nn3)=limn→∞n2⋅(1n3+nn3)=limn→∞n2⋅n+1n3=limn→∞n+12n2=0
\begin{align*}
&\lim_{n\to \infty}{\left(\frac{1}{n^3}+\frac{2}{n^3}+\cdots+\frac{n}{n^3}\right)}\\
=&\lim_{n\to \infty}{\frac{n}{2}\cdot \left(\frac{1}{n^3}+\frac{n}{n^3}\right)}\\
=&\lim_{n\to \infty}{\frac{n}{2}\cdot \frac{n+1}{n^3}}\\
=&\lim_{n\to \infty}{\frac{n+1}{2n^2}}\\
=&0\\
\end{align*}
====n→∞lim(n31+n32+⋯+n3n)n→∞lim2n⋅(n31+n3n)n→∞lim2n⋅n3n+1n→∞lim2n2n+10
limn→∞(1+12+122+⋯+12n−1)=limn→∞1×(1−(12)n−1)1−12=limn→∞2(1−(12)n−1)=2
\begin{align*}
&\lim_{n\to \infty}{\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}}\right)}\\
=&\lim_{n\to \infty}{\frac{1\times\left(1-(\frac{1}{2})^{n-1}\right)}{1-\frac{1}{2}}}\\
=&\lim_{n\to \infty}{2\left(1-(\frac{1}{2})^{n-1}\right)}\\
=&2
\end{align*}
===n→∞lim(1+21+221+⋯+2n−11)n→∞lim1−211×(1−(21)n−1)n→∞lim2(1−(21)n−1)2
有理化方案
limx→+∞x2+x+1−x2−x+1=limx→+∞(x2+x+1−x2−x+1)(x2+x+1+x2−x+1)x2+x+1+x2−x+1=limx→+∞x2+x+1−x2+x−1x2+x+1+x2−x+1=limx→+∞2xx2+x+1+x2−x+1=limx→+∞2x−2(x2+x+1+x2−x+1)=limx→+∞2(1+1x+1x2+1−1x+1x2)=limx→+∞2(1+0+0+1−0+0)=1 \begin{align*} &\lim_{x\to +\infty}{\sqrt{x^2+x+1}-\sqrt{x^2-x+1}}\\ =&\lim_{x\to +\infty}{\frac{(\sqrt{x^2+x+1}-\sqrt{x^2-x+1})(\sqrt{x^2+x+1}+\sqrt{x^2-x+1})}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}}\\ =&\lim_{x\to +\infty}{\frac{x^2+x+1-x^2+x-1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}}\\ =&\lim_{x\to +\infty}{\frac{2x}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}}\\ =&\lim_{x\to +\infty}{\frac{2}{\sqrt{x^{-2}}(\sqrt{x^2+x+1}+\sqrt{x^2-x+1})}}\\ =&\lim_{x\to +\infty}{\frac{2}{(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1-\frac{1}{x}+\frac{1}{x^2}})}}\\ =&\lim_{x\to +\infty}{\frac{2}{(\sqrt{1+0+0}+\sqrt{1-0+0})}}\\ =&1\\ \end{align*} =======x→+∞limx2+x+1−x2−x+1x→+∞limx2+x+1+x2−x+1(x2+x+1−x2−x+1)(x2+x+1+x2−x+1)x→+∞limx2+x+1+x2−x+1x2+x+1−x2+x−1x→+∞limx2+x+1+x2−x+12xx→+∞limx−2(x2+x+1+x2−x+1)2x→+∞lim(1+x1+x21+1−x1+x21)2x→+∞lim(1+0+0+1−0+0)21
limn→∞n(n+1−n)=limn→∞n(n+1−n)(n+1+n)n+1+n=limn→∞nn+1+n=limn→∞nn+1+n=limn→∞1n+1n+1=12 \begin{align*} &\lim_{n\to \infty}{\sqrt{n}}(\sqrt{n+1}-\sqrt{n})\\ =&\lim_{n\to \infty}{\frac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}}\\ =&\lim_{n\to \infty}{\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}}\\ =&\lim_{n\to \infty}{\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}}\\ =&\lim_{n\to \infty}{\frac{1}{\sqrt{\frac{n+1}{n}}+1}}\\ =&\frac{1}{2} \end{align*} =====n→∞limn(n+1−n)n→∞limn+1+nn(n+1−n)(n+1+n)n→∞limn+1+nnn→∞limn+1+nnn→∞limnn+1+1121
limx→−∞x2+x+1−x2−x+1=limx→−∞x2+x+1−x2+x−1x2+x+1+x2−x+1=limx→−∞2xx2+x+1+x2−x+1=limx→−∞2−1+1x1x2+−1−1x+1x2=−1 \begin{align*} &\lim_{x\to -\infty}{\sqrt{x^2+x+1}-\sqrt{x^2-x+1}}\\ =&\lim_{x\to -\infty}{\frac{x^2+x+1-x^2+x-1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}}\\ =&\lim_{x\to -\infty}{\frac{2x}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}}\\ =&\lim_{x\to -\infty}{\frac{2}{-\sqrt{1+\frac{1}{x}\frac{1}{x^2}}+-\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}}\\ =&-1 \end{align*} ====x→−∞limx2+x+1−x2−x+1x→−∞limx2+x+1+x2−x+1x2+x+1−x2+x−1x→−∞limx2+x+1+x2−x+12xx→−∞lim−1+x1x21+−1−x1+x212−1
由于x→−∞x\to -\inftyx→−∞,故而x=−x2≠x2x=-\sqrt{x^2}\ne\sqrt{x^2}x=−x2=x2
核心目的是在x→+∞x\to +\inftyx→+∞,时构建1x\frac{1}{x}x1形式的无穷小量;构建平方差是一个去除根号的好方法; 看到带根号的式子可以考虑这个方案;