函数极限的计算

极限存在的条件:

对于自变量趋向无穷大的情况:
lim⁡x→∞f(x)=A⇔=lim⁡x→+∞f(x)=lim⁡x→−∞f(x)=A\lim\limits_{x\rightarrow \infty}f(x)=A \Leftrightarrow = \lim\limits_{x\rightarrow +\infty}f(x)=\lim\limits_{x\rightarrow -\infty}f(x)=A

对于自变量趋向某一定点的情况:
lim⁡x→x0f(x)=A⇔=lim⁡x→x0+f(x)=lim⁡x→x0−f(x)=A\lim\limits_{x\rightarrow x_0}f(x)=A \Leftrightarrow = \lim\limits_{x\rightarrow x_0^+}f(x)=\lim\limits_{x\rightarrow x_0^-}f(x)=A

需要注意的函数:

  1. lim⁡x→∞ex\lim\limits_{x\rightarrow \infty}e^x

    1. lim⁡x→−∞ex=0\lim\limits_{x\rightarrow -\infty}e^x = 0
    2. lim⁡x→+∞ex=∞\lim\limits_{x\rightarrow +\infty}e^x=\infty

  2. lim⁡x→0e1x\lim\limits_{x\rightarrow 0}e^{\frac{1}{x}}

    1. lim⁡x→0+e1x=∞\lim\limits_{x\rightarrow 0^+}e^{\frac{1}{x}} = \infty
    2. lim⁡x→0−e1x=0\lim\limits_{x\rightarrow 0^-}e^{\frac{1}{x}}=0

  3. lim⁡x→∞arctan⁡x\lim\limits_{x\rightarrow \infty}\arctan x

    1. lim⁡x→+∞arctan⁡x=π2\lim\limits_{x\rightarrow +\infty}\arctan x=\frac{\pi}{2}
    2. lim⁡x→−∞arctan⁡x=−π2\lim\limits_{x\rightarrow -\infty}\arctan x = -\frac{\pi}{2}

  4. lim⁡x→0arctan⁡1x\lim\limits_{x\rightarrow 0}\arctan \frac{1}{x}

    1. lim⁡x→0+arctan⁡1x=π2\lim\limits_{x\rightarrow 0^+}\arctan \frac{1}{x} = \frac{\pi}{2}
    2. lim⁡x→0−arctan⁡1x=−π2\lim\limits_{x\rightarrow 0^-}\arctan \frac{1}{x} = -\frac{\pi}{2}

  5. 向下取整函数lim⁡x→0[x]\lim\limits_{x\rightarrow 0}[x]

    1. lim⁡x→0+[x]=0\lim\limits_{x\rightarrow 0^+}[x]=0
    2. lim⁡x→0−[x]=−1\lim\limits_{x\rightarrow 0^-}[x]=-1

  6. 分段函数的分段点.

极限运算法则:

∃±∃=∃\exists \pm \exists = \exists
̸∃±̸∃=undefined\not\exists \pm \not\exists = undefined
∃±̸∃≠∃\exists \pm \not\exists = \not\exists

极限计算

  1. 对于00,∞∞,0⋅∞\frac{0}{0}, \frac{\infty}{\infty},0\cdot\infty型极限式:

    1. 00\frac{0}{0} 型使用泰勒公式,等价无穷小和洛必达法则进行计算
    2. ∞∞\frac{\infty}{\infty} 型使用洛必达法则, 无穷大之间的比较等进行计算
    3. 0⋅∞0\cdot\infty 型转化成前两种进行计算(将简单的一个放到分母上)
      1. 00=01∞\frac{0}{0}=\frac{0}{\frac{1}{\infty}}
      2. ∞∞=∞10\frac{\infty}{\infty}=\frac{\infty}{\frac{1}{0}}

  2. 对于1∞,∞0,001^{\infty}, \infty^0, 0^0(0表示无穷小,并非真正的0)幂指函数:f(x)g(x)=eg(x)ln⁡f(x)=eln⁡f(x)1g(x)f(x)^{g(x)} = e^{g(x)\ln f(x)} = e^{\frac{\ln f(x)}{\frac{1}{g(x)}}}

    1. 都可以转化为:eAe^A
      1. 1∞=e∞⋅ln⁡1⇒A=lim⁡x→∗∞⋅ln⁡1=lim⁡x→∗∞⋅0=lim⁡x→∗01∞1^{\infty} = e^{\infty\cdot\ln1} \Rightarrow A=\lim\limits_{x\rightarrow*}\infty\cdot\ln1=\lim\limits_{x\rightarrow*}\infty\cdot0=\lim\limits_{x\rightarrow*}\frac{0}{\frac{1}{\infty}} (或者lim⁡x→∗∞10\lim\limits_{x\rightarrow*}\frac{\infty}{\frac{1}{0}})
      2. ∞0=e0⋅ln⁡∞⇒A=lim⁡x→∗0⋅ln⁡∞=lim⁡x→∗0⋅∞=lim⁡x→∗01∞\infty^0=e^{0\cdot \ln\infty} \Rightarrow A=\lim\limits_{x\rightarrow*}0\cdot\ln\infty = \lim\limits_{x\rightarrow*}0\cdot\infty = \lim\limits_{x\rightarrow*}\frac{0}{\frac{1}{\infty}} (或者lim⁡x→∗∞10\lim\limits_{x\rightarrow*}\frac{\infty}{\frac{1}{0}})
      3. 00=e0⋅ln⁡0⇒A=lim⁡x→∗0⋅ln⁡0=lim⁡x→∗0⋅∞=lim⁡x→∗01∞0^0 = e^{0\cdot\ln 0} \Rightarrow A=\lim\limits_{x\rightarrow*} 0\cdot\ln 0 = \lim\limits_{x\rightarrow*}0\cdot\infty = \lim\limits_{x\rightarrow*}\frac{0}{\frac{1}{\infty}} (或者lim⁡x→∗∞10\lim\limits_{x\rightarrow*}\frac{\infty}{\frac{1}{0}})
      4. 所以最终还是变成了00\frac{0}{0}或∞∞\frac{\infty}{\infty}类型的计算
    2. 基本公式
      1. lim⁡x→0(1+x)1x=e\lim\limits_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e
      2. lim⁡x→∞(1+1x)x=e\lim\limits_{x\rightarrow \infty}(1+\frac{1}{x})^x=e
    3. 扩展
      1. lim⁡x→∗(1+f(x))g(x)=lim⁡x→∗(1+f(x))1f(x)f(x)g(x)=eA\lim\limits_{x\rightarrow *}(1+f(x))^{g(x)}=\lim\limits_{x\rightarrow *}(1+f(x))^{\frac{1}{f(x)} f(x)g(x)}=e^A,其中A=lim⁡x→∗f(x)g(x)A=\lim\limits_{x\rightarrow *}f(x)g(x)

        eg1:lim⁡x→0(1−x)1x=e−1(A=−1)\lim\limits_{x\rightarrow0}(1-x)^{\frac{1}{x}} = e^{-1}(A=-1)
        eg2:lim⁡x→0(1+2x)1x=e2(A=2)\lim\limits_{x\rightarrow0}(1+2x)^{\frac{1}{x}} = e^2(A=2)
        eg3:lim⁡x→0(1−1x)kx=e−k(A=−k)\lim\limits_{x\rightarrow0}(1-\frac{1}{x})^{kx} = e^{-k}(A=-k)

      2. lim⁡x→0f(x)g(x)=lim⁡x→0(1+f(x)−1)g(x)=eA\lim\limits_{x\rightarrow0}f(x)^{g(x)}=\lim\limits_{x\rightarrow0}(1+f(x)-1)^{g(x)}=e^A,其中A=lim⁡x→∗(f(x)−1)g(x)A=\lim\limits_{x\rightarrow *}(f(x)-1)g(x)

        eg1: lim⁡x→∞(x+3x+6)x−1x\lim\limits_{x\rightarrow \infty}(\frac{x+3}{x+6})^{\frac{x-1}{x}},首先确定是1∞1^{\infty}类型, 然后计算A=lim⁡x→∞(x+3x+6−1)x−12=lim⁡x→∞x+3−x−6x+6⋅x−12=−32A=\lim\limits_{x\rightarrow \infty}(\frac{x+3}{x+6}-1)\frac{x-1}{2}=\lim\limits_{x\rightarrow \infty}\frac{x+3-x-6}{x+6}\cdot\frac{x-1}{2}=-\frac{3}{2},故答案为e−32e^{-\frac{3}{2}}
        eg2: lim⁡x→0cos⁡x1ln⁡(1+x2)\lim\limits_{x\rightarrow0}{\cos x}^{\frac{1}{\ln(1+x^2)}}, 首先确定为1∞1^{\infty}类型,然后计算A=lim⁡x→0cos⁡x−1ln⁡(1+x2)=lim⁡x→0−12x2x2=−12A=\lim\limits_{x\rightarrow0}\frac{\cos x-1}{\ln(1+x^2)}=\lim\limits_{x\rightarrow0}\frac{-\frac{1}{2}x^2}{x^2} =-\frac{1}{2},故答案为e−12e^{-\frac{1}{2}}

      3. lim⁡x→0(1+2x2)1x\lim\limits_{x\rightarrow0}(\frac{1+2^x}{2})^{\frac{1}{x}},为1∞1^{\infty}类型
        A=lim⁡x→0(1+2x2−1)1x=lim⁡x→02x−12x=lim⁡x→02xln⁡22=12ln⁡2=ln⁡2A=\lim\limits_{x\rightarrow0}(\frac{1+2^x}{2}-1)\frac{1}{x}=\lim\limits_{x\rightarrow0}\frac{2^x-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2^x\ln2}{2}=\frac{1}{2}\ln2=\ln\sqrt{2}
        ∴lim⁡x→0(1+2x2)1x=eln⁡2=2\therefore\lim\limits_{x\rightarrow0}(\frac{1+2^x}{2})^{\frac{1}{x}}=e^{\ln\sqrt{2}}=\sqrt2
        同理可得
        lim⁡x→0(1+axb)1x=e1aln⁡b=ba\lim\limits_{x\rightarrow0}(\frac{1+a^x}{b})^{\frac{1}{x}}=e^{\frac{1}{a}\ln b}=\sqrt[a]{b}
        lim⁡x→0(ax+bx2)1x=e12ln⁡ab=ab\lim\limits_{x\rightarrow0}(\frac{a^x+b^x}{2})^{\frac{1}{x}}=e^{\frac{1}{2}\ln ab}=\sqrt{ab}

  3. 对于∞−∞\infty - \infty类型未定式(通过下列三种方式转换为00\frac{0}{0}或者∞∞\frac{\infty}{\infty}再进行计算):

    1. 通分

      eg1:
      lim⁡x→0(1sin⁡2x−cos⁡2xx2)=lim⁡x→0x2−sin⁡2xcos⁡2xx2sin⁡x=lim⁡x→0x2−14sin⁡22xx4=lim⁡x→02x−14⋅2sin⁡2xcos⁡2x⋅24x3\lim\limits_{x\rightarrow0}(\frac{1}{\sin^2x}-\frac{\cos^2x}{x^2})=\lim\limits_{x\rightarrow0}\frac{x^2-\sin^2x\cos^2x}{x^2\sin x}=\lim\limits_{x\rightarrow0}\frac{x^2-\frac{1}{4}\sin^22x}{x^4}=\lim\limits_{x\rightarrow0}\frac{2x-\frac{1}{4}\cdot2\sin2x\cos2x\cdot2}{4x^3}
      =lim⁡x→02x−12sin⁡4x4x3=lim⁡x→02x−12[4x−16(4x)3+o(x3)]4x3=lim⁡x→04x33x3=43=\lim\limits_{x\rightarrow0}\frac{2x-\frac{1}{2}\sin4x}{4x^3}=\lim\limits_{x\rightarrow0}\frac{2x-\frac{1}{2}[4x-\frac{1}{6}(4x)^3+o(x^3)]}{4x^3}=\lim\limits_{x\rightarrow0}\frac{4x^3}{3x^3}=\frac{4}{3}
      eg2:
      lim⁡x→0(1sin⁡2x−1x2)=lim⁡x→0x2−sin⁡2xx2sin⁡2x=lim⁡x→0(x+sin⁡x)(x−sin⁡x)x4=lim⁡x→02x⋅16x3x4=13\lim\limits_{x\rightarrow0}(\frac{1}{\sin^2x}-\frac{1}{x^2})=\lim\limits_{x\rightarrow0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\lim\limits_{x\rightarrow0}\frac{(x+\sin x)(x-\sin x)}{x^4}=\lim\limits_{x\rightarrow0}\frac{2x\cdot \frac{1}{6}x^3}{x^4}=\frac{1}{3}

    2. 倒代换

      lim⁡x→+∞[x2(e1x−1)−x]\lim\limits_{x\rightarrow +\infty}[x^2(e^{\frac{1}{x}}-1)-x]
      =lim⁡t→0+et+1t2−1t=\lim\limits_{t\rightarrow 0^+}\frac{e^t+1}{t^2} - \frac{1}{t}
      =lim⁡t→0+et+1−t2t=\lim\limits_{t\rightarrow 0^+}\frac{e^t+1-t}{2t}
      =lim⁡t→0+et−12t=\lim\limits_{t\rightarrow 0^+}\frac{e^t-1}{2t}
      =lim⁡t→0+t2t=12=\lim\limits_{t\rightarrow 0^+}\frac{t}{2t} = \frac{1}{2}

    3. 分子有理化

      eg1:
      lim⁡x→01+tan⁡x−1−sin⁡xx1+sin⁡2x−x=lim⁡x→01+tan⁡x−1+sin⁡xx1+sin⁡2x−x11+tan⁡x+1−sin⁡x=12lim⁡x→0tan⁡x+sin⁡xx(1+sin⁡2x−1)\lim\limits_{x\rightarrow 0}\frac{\sqrt{1+\tan x}- \sqrt{1-\sin x}}{x\sqrt{1+\sin^2 x-x}}=\lim\limits_{x\rightarrow 0}\frac{1+\tan x - 1 + \sin x}{x\sqrt{1+\sin^2 x -x}}\frac{1}{\sqrt{1+\tan x} + \sqrt{1-\sin x}}=\frac{1}{2} \lim\limits_{x\rightarrow 0}\frac{\tan x + \sin x}{x(\sqrt{1+\sin^2 x}-1)}
      =12lim⁡x→0tan⁡x(1−cos⁡x)x.12sin⁡2x=12lim⁡x→0x12x2x12x2=12=\frac{1}{2} \lim\limits_{x\rightarrow 0}\frac{\tan x(1 - \cos x)}{x.\frac{1}{2}\sin^2 x}=\frac{1}{2} \lim\limits_{x\rightarrow 0}\frac{x\frac{1}{2}x^2}{x\frac{1}{2}x^2} = \frac{1}{2}