洛必达法则训练题

Cite: 洛必达法则: https://www.cnblogs.com/Preparing/p/17065708.html

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Exercise

first

若条件符合,洛必达法则可连续复用,直至求出极限为止

\[\begin{eqnarray} \lim_{x \to 0} \frac{x-\sin{x}}{x^{3}}=? \\ \\ \lim_{x \to 0} \frac{x-\sin{x}}{x^{3}}=\lim_{x \to 0} \frac{(x)'-(\sin{x})'}{(x^{3})'}=\lim_{x \to 0} \frac{1-\cos{x}}{3x^{2}} \\ \\ \lim_{x \to 0} \frac{1-\cos{x}}{3x^{2}}=\lim_{x \to 0} \frac{(1)'-(\cos{x})'}{(3x^{2})'}=\lim_{x \to 0} \frac{0-(-\sin{x})}{6\cdot (x^{2})'} \\ \\ =\lim_{x \to 0} \frac{\sin{x}}{6x} =\frac{1}{6} \cdot \lim_{x \to 0}\frac{\sin{x}}{x} =\frac{1}{6} \\ \\ \therefore \lim_{x \to 0} \frac{x-\sin{x}}{x^{3}}=\frac{1}{6} \end{eqnarray} \]

second

洛必达法则可以仅应用在解题的某一步或某几步,不必全程使用

\[\begin{eqnarray} \lim_{x \to 1} \frac{\ln{x}}{(x-1)^{2}}=? \\ \\ \lim_{x \to 1} \frac{\ln{x}}{(x-1)^{2}}=\lim_{x \to 1} \frac{(\ln{x})'}{[(x-1)^{2}]'} \\ \\ (\ln{x})'=\frac{1}{x}, \quad [(x-1)^{2}]'=2(x-1)\cdot 1=2(x-1) \\ \\ \lim_{x \to 1} \frac{(\ln{x})'}{[(x-1)^{2}]'}=\lim_{x \to 1} \frac{\frac{1}{x}}{2(x-1)} \\ \\ \because \lim_{x \to 1} \frac{\frac{1}{x}}{2(x-1)} 非\frac{0}{0}型亦非\frac{\infty}{\infty}型,不适于洛必达法则,洛必达法则遂避之 \\ \\ \lim_{x \to 1} \frac{\frac{1}{x}}{2(x-1)}=\frac{1}{2} \lim_{x \to 1} \frac{1}{x(x-1)} \\ \\ \because \lim_{x \to 1} x(x-1)=0, \quad \therefore x(x-1)为 x \to 1 之无穷小 \\ \\ \therefore \lim_{x \to 1} \frac{1}{x(x-1)}=\infty \quad (\infty :无穷大) \\ \\ \frac{1}{2} \lim_{x \to 1} \frac{1}{x(x-1)}=\frac{1}{2} \cdot \infty =\infty \\ \\ \lim_{x \to 1} \frac{\ln{x}}{(x-1)^{2}}=\infty \end{eqnarray} \]

third

除\(\frac{0}{0} 和 \frac{\infty}{\infty}\) 这两种未定式外,

还有 \(0\cdot \infty,\infty-\infty,0^{0},\infty^{0},1^{\infty}\) 型的未定式,

它们经过适当变形后可以转化为 \(\frac{0}{0} 和 \frac{\infty}{\infty}\) 这两种基本未定式

\[\begin{eqnarray} \lim_{x \to 0^{+}} x\ln{x}=? \\ \\ \lim_{x \to 0^{+}} x=0, \quad \lim_{x \to 0^{+}} \ln{x}= \infty \\ \\ \lim_{x \to 0^{+}} x\ln{x}为 \frac{0}{\infty }型未定式, 不适于洛必达法则, 需变形 \\ \\ 设\ln{x}=t, \quad x\ln{x}=xt=\frac{t}{\frac{1}{x}}=\frac{\ln{x}}{\frac{1}{x}} \\ \\ \lim_{x \to 0^{+}} x\ln{x}= \lim_{x \to 0^{+}} \frac{\ln{x}}{\frac{1}{x}} \\ \\ \lim_{x \to 0^{+}} \frac{1}{x}= +\infty, \quad \lim_{x \to 0^{+}} \ln{x}=o (无穷小) \\ \\ \because \lim_{x \to 0^{+}} \frac{\ln{x}}{\frac{1}{x}} 为\frac{\infty }{\infty }型未定式, 适于洛必达法则 \\ \\ \therefore \lim_{x \to 0^{+}} \frac{(\ln{x})'}{(\frac{1}{x})'} = \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{-x^{-2}} =\lim_{x \to 0^{+}} -1\cdot \frac{x^{-1}}{x^{-2}} \\ \\ =\lim_{x \to 0^{+}} -x =0 \\ \\ \lim_{x \to 0^{+}} x\ln{x}=0 \end{eqnarray} \]

fourth

\[\begin{eqnarray} \lim_{x \to \frac{\pi}{2}} (\sec{x}-\tan{x})=? \\ \\ \because \lim_{x \to \frac{\pi}{2}} \sec{x}=\infty, \quad \lim_{x \to \frac{\pi}{2}} \tan{x}=\infty \\ \\ \therefore 原式为\infty - \infty 型未定式,不适于洛必达法则 \\ \\ 遂变形: \quad \sec{x}=\frac{1}{\cos{x}}, \quad \tan{x}=\frac{\sin{x}}{\cos{x}} \\ \\ \sec{x}-\tan{x}=\frac{1-\sin{x}}{\cos{x}} \\ \\ \because \lim_{x \to \frac{\pi}{2}} (1-\sin{x})=1-1=0, \quad \lim_{x \to \frac{\pi}{2}} \cos{x}=0 \\ \\ \therefore \lim_{x \to \frac{\pi}{2}} \frac{1-\sin{x}}{\cos{x}} 为\frac{0}{0}型未定式,适于洛必达法则 \\ \\ \lim_{x \to \frac{\pi}{2}} \frac{(1-\sin{x})'}{(\cos{x})'}= \lim_{x \to \frac{\pi}{2}} \frac{-\cos{x}}{-\sin{x}} \\ \\ \because \lim_{x \to \frac{\pi}{2}} -\cos{x}=0, \quad \lim_{x \to \frac{\pi}{2}} -\sin{x}=-1 \\ \\ \therefore \lim_{x \to \frac{\pi}{2}} \frac{-\cos{x}}{-\sin{x}}=0 \\ \\ \lim_{x \to \frac{\pi}{2}} (\sec{x}-\tan{x})=0 \end{eqnarray} \]

fifth

\[\begin{eqnarray} \lim_{x \to 0^{+}} x^{x}=? \\ \\ x^{x}转成以e为底的幂指函数: \quad x^{x}=e^{x\ln{x}} \\ \\ \lim_{x \to 0^{+}} x^{x}= \lim_{x \to 0^{+}} e^{x\ln{x}} \\ \\ 设\lim_{x \to 0^{+}} x\ln{x}=\gamma \\ \\ \lim_{x \to 0^{+}} e^{x\ln{x}} \Rightarrow e^ {\gamma } \\ \\ \because \lim_{x \to 0^{+}} x=0, \quad \lim_{x \to 0^{+}} \ln{x} =\infty \\ \\ \therefore \gamma=0 \\ \\ e^ { \gamma}=e^{0}=1 \\ \\ \lim_{x \to 0^{+}} x^{x}=1 \end{eqnarray} \]

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