高数 03.02洛必达法则

第三章第二节洛必达法则 \color{blue}{第三章第二节洛必达法则}

一、00 型未定式一、\dfrac{0}{0}型未定式
二、∞∞ 型未定式二、\dfrac{\infty}{\infty}型未定式
三、其他未定式三、其他未定式

本节研究：函数之商的极限limf(x)g(x) (00 或∞∞ 型) 本节研究：\\ 函数之商的极限\lim{\dfrac{f(x)}{g(x)}}(\dfrac{0}{0}或\dfrac{\infty}{\infty}型)
转化↓洛必达法则转化 \downarrow 洛必达法则
导数之商的极限limf ′ (x)g ′ (x) 导数之商的极限\lim{\dfrac{f^{\prime}(x)}{g^{\prime}(x)}}

一、00 型未定式 \color{blue}{一、\dfrac{0}{0}型未定式}

定理1.1)lim x→a f(x)=lim x→a F(x)=02)f(x)与F(x)在U ° (a)内可导,且F ′ (x)≠03)lim x→a f ′ (x)F ′ (x) 存在(或为∞)⟹lim x→a f(x)F(x) =lim x→a f ′ (x)F ′ (x) (洛必达法则) 定理1.\\ 1)\lim_{x \rightarrow a}{f(x)} = \lim_{x \rightarrow a}{F(x)} = 0 \\ 2)f(x)与F(x)在\mathring{U}(a)内可导,且F^{\prime}(x) \neq 0 \\ 3)\lim_{x \rightarrow a}{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}}存在(或为\infty)\\ \Longrightarrow \lim_{x \rightarrow a}{\dfrac{f(x)}{F(x)} } = \lim_{x \rightarrow a}{\dfrac{f^{\prime}(x)}{F^{\prime}(x)} } \quad (洛必达法则)
该定理的证明用到了柯西中值定理该定理的证明用到了柯西中值定理

洛必达法则：lim x→a f(x)g(x) =lim x→a f ′ (x)g ′ (x) \lim_{x \rightarrow a}{\dfrac{f(x)}{g(x)}} = \lim_{x \rightarrow a}{\dfrac{f^{\prime}(x)}{g^{\prime}(x)}}

推论1.定理1中x→a换为x→a + ,x→a − ,x→∞,x→+∞,x→−∞之一,2)作相应的修改，定理1仍然成立. 推论1.定理1中x \rightarrow a换为\\ x \rightarrow a^+, x \rightarrow a^-, x \rightarrow \infty, x \rightarrow +\infty, x \rightarrow -\infty之一,\\ 2)作相应的修改，定理1仍然成立.
推论2.若limf ′ (x)F ′ (x) 仍属00 型,且f ′ (x),F ′ (x)满足定理1条件,则limf(x)F(x) =limf ′ (x)F ′ (x) =limf ′′ (x)F ′′ (x) 推论2.若\lim{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}}仍属\dfrac{0}{0}型,且f^{\prime}(x),F^{\prime}(x)满足定理1条件,则\\ \lim{\dfrac{f(x)}{F(x)}} = \lim{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}} = \lim{\dfrac{f^{\prime \prime}(x)}{F^{\prime \prime}(x)}}

例1.求lim x→1 x 3 −3x+2x 3 −x 2 −x+1 例1.求\lim_{x \rightarrow 1}{\dfrac{x^3 - 3x +2}{x^3 - x^2 - x + 1}}
解：原式=lim x→1 3x 2 −33x 2 −2x−1 =lim x→1 6x6x−2 =lim x→1 66−2 =32 解：原式\\ = \lim_{x \rightarrow 1}{\dfrac{3x^2 - 3}{3x^2 - 2x - 1}} \\ = \lim_{x \rightarrow 1}{\dfrac{6x}{6x - 2}} \\ = \lim_{x \rightarrow 1}{\dfrac{6}{6 - 2}} \\ = \dfrac{3}{2}
注意：以上前两步求导是因为满足00 型，满足洛必达法则。不是未定式不能用洛必达法则！注意：以上前两步求导是因为满足\dfrac{0}{0}型，满足洛必达法则。不是未定式不能用洛必达法则！

例2.求lim x→+∞ π2 −arctanx1x 例2.求\lim_{x \rightarrow +\infty}{\dfrac{\frac{\pi}{2} - \arctan{x}}{\frac{1}{x}}}
解：原式=lim x→+∞ 0−11+x 2 −1x 2 =lim x→+∞ x 2 1+x 2 =lim x→+∞ 11x 2 +1 =1 解：\\ 原式= \lim_{x \rightarrow +\infty}{\dfrac{0 - \dfrac{1}{1 + x^2}}{-\dfrac{1}{x^2}}} \\ = \lim_{x \rightarrow +\infty}{\dfrac{x^2}{1 + x^2}} \\ = \lim_{x \rightarrow +\infty}{\dfrac{1}{\frac{1}{x^2} + 1}} \\ = 1

二、∞∞ 型未定式 \color{blue}{二、\dfrac{\infty}{\infty}型未定式}

定理2.1)lim x→a |f(x)|=lim x→a |F(x)|=∞2)f(x)与F(x)在U ° (a)内可导,且F ′ (x)≠03)lim x→a f ′ (x)F ′ (x) 存在(或为∞)⟹lim x→a f(x)F(x) =lim x→a f ′ (x)F ′ (x) 定理2.\\ 1)\lim_{x \rightarrow a}{|f(x)|} = \lim_{x \rightarrow a}{|F(x)|} = \infty\\ 2)f(x)与F(x)在\mathring{U}(a)内可导,且F^{\prime}(x) \neq 0\\ 3)\lim_{x \rightarrow a}{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}}存在(或为\infty) \\ \Longrightarrow \lim_{x \rightarrow a}{\dfrac{f(x)}{F(x)}} = \lim_{x \rightarrow a}{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}}

例3.求lim x→+∞ lnxx n (n>0). 例3.求\lim_{x \rightarrow +\infty}{\dfrac{\ln{x}}{x^n}} (n > 0).
解：原式=lim x→+∞ 1x nx n−1 =lim x→+∞ 1nx n =0 解：原式 \\ =\lim_{x \rightarrow +\infty}{\dfrac{\dfrac{1}{x}}{nx^{n - 1}}} \\ =\lim_{x \rightarrow +\infty}{\dfrac{1}{nx^n}} \\ = 0

例4.求lim x→+∞ x n e λx (n>0,λ>0). 例4.求\lim_{x \rightarrow +\infty}{\dfrac{x^n}{e^{\lambda x}}} (n > 0, \lambda > 0).
解：1)n为正整数时:原式=lim x→+∞ nx n−1 λe λx =lim x→+∞ n!λ n e λx =n!λ n lim x→+∞ 1e λx =n!λ n ⋅0=02)n不是正整数时,存在正整数k，使当x>1时,x k <x n <x k+1 从而x k e λx <x n e λx <x k+1 e λx 由1)可知lim x→+∞ x k e λx =0lim x→+∞ x k+1 e λx =0根据夹逼定理得:lim x→+∞ x n e λx =0 解：\\ 1) n为正整数时: 原式= \lim_{x \rightarrow +\infty}{\dfrac{nx^{n - 1}}{\lambda e^{\lambda x}}} \\ = \lim_{x \rightarrow +\infty}{\dfrac{n!}{\lambda^n e^{\lambda x}}} \\ =\dfrac{n!}{\lambda^n} \lim_{x \rightarrow +\infty}{\dfrac{1}{e^{\lambda x}}} \\ =\dfrac{n!}{\lambda^n} \cdot 0 \\ = 0 \\ 2)n不是正整数时,\\ 存在正整数k，使当x > 1时, \\ x^k

说明：1)例3,例4表明x→+∞时,lnx,x n (n>0),e λx (λ>0)后者比前者趋于+∞更快.2)在满足定理条件的某些情况下洛必达法则不能解决所有计算问题.例如，lim x→+∞ 1+x 2 − − − − − √ x =lim x→+∞ x1+x 2 − − − − − √ 3)若limf ′ (x)F ′ (x) 不存在(≠∞)时,limf(x)F(x) ≠limf ′ (x)F ′ (x) 例如，lim x→+∞ x+sinxx ≠lim x→+∞ 1+cosx1 lim x→+∞ 1+cosx1 极限不存在说明：\\ 1)例3,例4表明 x \rightarrow +\infty时, \\ \ln{x}, x^n(n > 0), e^{\lambda x}(\lambda >0) \\ 后者比前者趋于+\infty更快.\\ 2)在满足定理条件的某些情况下洛必达法则不能解决所有计算问题.\\ 例如，\lim_{x \rightarrow +\infty}{\dfrac{\sqrt{1 + x^2}}{x}} = \lim_{x \rightarrow +\infty}{\dfrac{x}{\sqrt{1 + x^2}}} \\ 3)若\lim{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}}不存在(\neq \infty)时,\\ \lim{\dfrac{f(x)}{F(x)}} \neq \lim{\dfrac{f^{\prime}(x)}{F^{\prime}(x)}}\\ 例如，\lim_{x \rightarrow +\infty}{\dfrac{x + \sin{x}}{x}} \neq \lim_{x \rightarrow +\infty}{\dfrac{1 + \cos{x}}{1}} \\ \lim_{x \rightarrow +\infty}{\dfrac{1 + \cos{x}}{1}}极限不存在

三、其他未定式：0⋅∞,∞−∞,0 0 ,1 ∞ ,∞ 0 型 \color{blue}{三、其他未定式：0\cdot \infty, \infty - \infty, 0^0, 1^{\infty}, \infty^{0}型}

解决方法：通分、去倒数、取对数解决方法：通分、去倒数、取对数
例5.求lim x→0 + x n lnx(n>0). 例5.求\lim_{x \rightarrow 0^+}{x^n \ln{x}} (n > 0).
解：lim x→0 + x n lnx=lim x→0 + lnxx −n =lim x→0 + 1x −nx −n−1 =lim x→0 + 1−nx −n =−1n lim x→0 + x n =0 解：\\ \lim_{x \rightarrow 0^+}{x^n \ln{x}} \\ = \lim_{x \rightarrow 0^+}{\dfrac{\ln{x}}{x^{-n}}} \\ = \lim_{x \rightarrow 0^+}{\dfrac{\dfrac{1}{x}}{-nx^{-n-1}}} \\ = \lim_{x \rightarrow 0^+}{\dfrac{1}{-nx^{-n}}} \\ = -\dfrac{1}{n}\lim_{x \rightarrow 0^+}{x^n} \\ = 0

例6.求lim x→π2 (secx−tanx). 例6.求\lim_{x \rightarrow \frac{\pi}{2}}{(\sec{x} - \tan{x})}.
解：lim x→π2 (secx−tanx)=lim x→π2 1−sinxcosx =lim x→π2 −cosx−sinx =0 解：\\ \lim_{x \rightarrow \frac{\pi}{2}}{(\sec{x} - \tan{x})} \\ = \lim_{x \rightarrow \frac{\pi}{2}}{\dfrac{1 - \sin{x}}{\cos{x}}} \\ = \lim_{x \rightarrow \frac{\pi}{2}}{\dfrac{- \cos{x}}{-\sin{x}}} \\ = 0

例7.求lim x→0 + x x . 例7.求\lim_{x \rightarrow 0^+}{x^x}.
解：lim x→0 + x x =lim x→0 + e xlnx =e (lim x→0 + xlnx) =e (lim x→0 + lnx1x ) =e (lim x→0 + 1x −1x 2 ) =e (lim x→0 + −x) =e 0 =1 解：\\ \lim_{x \rightarrow 0^+}{x^x} \\ = \lim_{x \rightarrow 0^+}{e^{x \ln{x}}} \\ = e^{(\lim_{x \rightarrow 0^+}{x \ln{x}})} \\ = e^{(\lim_{x \rightarrow 0^+}{\frac{\ln{x}}{\frac{1}{x}}})} \\ = e^{(\lim_{x \rightarrow 0^+}{\frac{\frac{1}{x}}{-\frac{1}{x^2}}})} \\ = e^{(\lim_{x \rightarrow 0^+}{-x})} \\ = e^0 \\ = 1

例8.求lim x→0 tanx−xx 2 sinx . 例8.求\lim_{x \rightarrow 0}{\dfrac{\tan{x} - x}{x^2 \sin{x}}}.
解：lim x→0 tanx−xx 2 sinx =lim x→0 1cos 2 x −12xsinx+x 2 cosx =lim x→0 sin 2 xx(2sinx+xcosx)cos 2 x =lim x→0 x 2 x(2x+xcosx)cos 2 x =lim x→0 1(2+cosx)cos 2 x =lim x→0 1(2+1)⋅1 2 =13 解：\\ \lim_{x \rightarrow 0}{\dfrac{\tan{x} - x}{x^2 \sin{x}}} \\ = \lim_{x \rightarrow 0}{\dfrac{\frac{1}{\cos^2{x}} - 1}{2x \sin{x} + x^2 \cos{x}}} \\ = \lim_{x \rightarrow 0}{\dfrac{\sin^2{x}}{x(2 \sin{x} + x \cos{x})\cos^2{x}}} \\ = \lim_{x \rightarrow 0}{\dfrac{x^2}{x(2 x + x \cos{x})\cos^2{x}}} \\ = \lim_{x \rightarrow 0}{\dfrac{1}{(2 + \cos{x})\cos^2{x}}} \\ = \lim_{x \rightarrow 0}{\dfrac{1}{(2 + 1) \cdot 1^2}} \\ = \dfrac{1}{3}
使用洛必达法则需要注意的几个问题：
1、使用洛必达法则之前，应该先检验其条件是否满足.
2、如果使用洛必达法则之后，命题仍是未定型极限，且仍符合洛必达法则的条件，可以再次使用洛必达法则.
3、如果00 \dfrac{0}{0}型或∞∞ \dfrac{\infty}{\infty}型极限中含有非零因子，该非零因子可以单独求极限，不必参与洛必达法则运算，以简化运算.
4、如果能进行等价无穷小代换或更等变形配合使用洛必达法则，也可以简化运算.