《高等数学》习题4.2选做

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用洛必达法则求下列极限：

  1. \displaystyle \lim _{x\rightarrow 0}\frac{2^{x} -1}{3^{x} -1}

解：\displaystyle =\lim _{x\rightarrow 0}\frac{2^{x}\ln 2}{3^{x}\ln 3} =\frac{\ln 2}{\ln 3}。

  1. \displaystyle \lim _{x\rightarrow 0}\frac{\cos x-1}{x-\ln( 1+x)}

解：\displaystyle =\lim _{x\rightarrow 0}\frac{-\sin x}{1-\frac{1}{1+x}} =\lim _{x\rightarrow 0}\frac{-\cos x}{( 1+x)^{-2}} =-1

  1. \displaystyle \lim _{x\rightarrow 0}\left(\frac{1}{\ln\left( x+\sqrt{1+x^{2}}\right)} -\frac{1}{\ln( 1+x)}\right)

解：

\begin{aligned} & =\lim _{x\rightarrow 0}\frac{\ln( 1+x) -\ln\left( x+\sqrt{1+x^{2}}\right)}{\ln\left( x+\sqrt{1+x^{2}}\right) \cdotp \ln( 1+x)}\\ & =\lim _{x\rightarrow 0}\frac{\frac{1}{1+x} -\frac{1}{\sqrt{1+x^{2}}}}{\frac{1}{1+x}\ln\left( x+\sqrt{1+x^{2}}\right) +\frac{1}{\sqrt{1+x^{2}}}\ln( 1+x)}\\ & =\lim _{x\rightarrow 0}\frac{\sqrt{1+x^{2}} -( 1+x)}{\sqrt{1+x^{2}}\ln\left( x+\sqrt{1+x^{2}}\right) +( 1+x)\ln( 1+x)}\\ & =\lim _{x\rightarrow 0}\frac{\frac{x}{\sqrt{1+x^{2}}} -1}{2+\frac{x}{\sqrt{1+x^{2}}}\ln\left( x+\sqrt{1+x^{2}}\right) +\ln( 1+x)}\\ & =-\frac{1}{2} \end{aligned}
  1. \displaystyle \lim _{x\rightarrow \frac{\pi }{2}}\frac{\tan 3x}{\tan x}

解：\displaystyle =\lim _{x\rightarrow \frac{\pi }{2}}\frac{3\sec^{2} 3x}{\sec^{2} x} =\lim _{x\rightarrow \frac{\pi }{2}}\frac{3\cos^{2} x}{\cos^{2} 3x}，而 \displaystyle \lim _{x\rightarrow \frac{\pi }{2}}\frac{\cos x}{\cos 3x} =\lim _{x\rightarrow \frac{\pi }{2}}\frac{\sin x}{3\sin 3x}，因此原式 \displaystyle =\frac{3}{9} =\frac{1}{3}。

  1. \displaystyle \lim _{x\rightarrow 0}\frac{\ln(\cos ax)}{\ln(\cos bx)}

解：\displaystyle =\lim _{x\rightarrow 0}\frac{a\tan ax}{b\tan bx} =\frac{a^{2}}{b^{2}}

  1. \displaystyle \lim _{x\rightarrow 0^{+}} x^{a}\ln x\quad ( a >0)

解：\displaystyle =\lim _{x\rightarrow 0^{+}}\frac{\ln x}{x^{-a}} =\lim _{x\rightarrow 0^{+}}\frac{1/x}{-ax^{-a-1}} =\lim _{x\rightarrow 0^{+}} -\frac{x^{a}}{a} =0

  1. \displaystyle \lim _{x\rightarrow \frac{\pi }{2}^{-}}(\tan x)^{2x-\pi }

解：

\begin{aligned} & =\exp\left[\lim _{x\rightarrow \pi /2^{-}}( 2x-\pi )\ln\tan x\right]\\ & =\exp\left[\lim _{x\rightarrow \pi /2^{-}}\frac{\sec^{2} x/\tan x}{-\frac{2}{( x-\pi /2)^{2}}}\right]\\ & =\exp\left[\lim _{x\rightarrow \pi /2^{-}}\frac{( x-\pi /2)^{2}}{-\sin 2x}\right]\\ & =\exp\left[\lim _{x\rightarrow \pi /2^{-}}\frac{2x-\pi }{-2\cos 2x}\right]\\ & =1 \end{aligned}
  1. \displaystyle \lim _{x\rightarrow \infty }\left( a^{1/x} -1\right) x\quad ( a >0)

解：\displaystyle =\lim _{x\rightarrow \infty }\frac{a^{1/x}\ln a\cdot \left( -1/x^{2}\right)}{-1/x^{2}} =\lim _{x\rightarrow \infty } a^{1/x}\ln a=\ln a

  1. \displaystyle \lim _{y\rightarrow 0}\frac{y-\arcsin y}{\sin^{3} y}

解：\displaystyle =\lim _{y\rightarrow 0}\frac{1-\frac{1}{\sqrt{1-y^{2}}}}{3y^{2}} =\frac{1}{3}\lim _{y\rightarrow 0}\frac{\sqrt{1-y^{2}} -1}{y^{2}} =\frac{1}{3}\lim _{y\rightarrow 0}\frac{\frac{-y}{\sqrt{1-y^{2}}}}{2y} =-\frac{1}{6}

  1. \displaystyle \lim _{x\rightarrow 1}\frac{x^{x} -x}{\ln x-x+1}

解：\displaystyle =\lim _{x\rightarrow 1}\frac{x^{x}( 1+\ln x) -1}{\frac{1}{x} -1} =\lim _{x\rightarrow 1}\frac{x^{x+1}( 1+\ln x) -x}{1-x} =-\lim _{x\rightarrow 1}\left[ x^{x}( x\ln x+x+1)( 1+\ln x) +x^{x} -1\right] =-2。

  1. \displaystyle \lim _{x\rightarrow +\infty }\left(\frac{2}{\pi }\arctan x\right)^{x}

解：

\begin{aligned} & =\exp\left[\lim _{x\rightarrow +\infty } x\ln\left(\frac{2}{\pi }\arctan x\right)\right]\\ & =\exp\left[\lim _{x\rightarrow +\infty }\frac{\frac{1}{\left( 1+x^{2}\right)\arctan x}}{-\frac{1}{x^{2}}}\right]\\ & =\exp\left[\lim _{x\rightarrow +\infty } -\frac{x^{2}}{\left( 1+x^{2}\right)\arctan x}\right]\\ & =\exp\left[\lim _{x\rightarrow +\infty } -\frac{2x^{2}}{\pi \left( 1+x^{2}\right)}\right]\\ & =\mathrm{e}^{-2/\pi } \end{aligned}