错题集-拉格朗日中值定理

发表: 4/26/2026 更新: 4/27/2026 字数: 0 字 时长: 0 分钟

① $f(b)-f(a)、\frac{f(b)-f(a)}{b-a}、f(a)\ne f(b)$ - L
② $f(a)、f(c)、f(b)\mathrm{或}{f}^{\prime }(a)、f^{\prime }(c)、f^{\prime }(b)$ - 2L

$\mathrm{例}1.\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=e.\underset{x\to \mathrm{\infty }}{lim}[f(x+2)-f(x)]=\underset{x\to \mathrm{\infty }}{lim}(\frac{x+a}{x-a}{)}^x,\mathrm{求}a.$

解:
$\frac{f(x+2)-f(x)}{2}=f^{\prime }(\zeta ),\zeta \in (x,x+2);$
$\mathrm{即}f(x+2)-f(x)=2f^{\prime }(\zeta )$
$\mathrm{左}=2\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(\zeta )=2e;$(夹逼定理)
$\mathrm{右}=\underset{x\to \mathrm{\infty }}{lim}[(1+\frac{2a}{x-a}{)}^{\frac{x-a}{2a}}{]}^{x\frac{2a}{x-a}}=e^{2a};$
$\Rightarrow e^{2a}=2e\Rightarrow 2a=1+\ln 2\Rightarrow a=\frac{1+\ln 2}{2}.$

$\mathrm{例}2.\mathrm{求}\underset{x\to \mathrm{\infty }}{lim}{x}^2(e^{\frac{1}{2x-1}}-e^{\frac{1}{2x+1}}).$

解:
$\mathrm{令}f(x)=e^{\frac{1}{2x}},f^{\prime }(x)=-\frac{e^{\frac{1}{2x}}}{2x^2}$
$e^{\frac{1}{2x-1}}-e^{\frac{1}{2x+1}}=-(f(x+\frac{1}{2})-f(x-\frac{1}{2}))=-f^{\prime }(\zeta ),\zeta \in (x-\frac{1}{2},x+\frac{1}{2})$
$\mathrm{得}{f}^{\prime }(\zeta )=\frac{e^{\frac{1}{2\zeta }}}{2{\zeta }^2}$
$\mathrm{原}\mathrm{式}=\underset{x\to \mathrm{\infty }}{lim}{x}^2\frac{e^{\frac{1}{2\zeta }}}{2{\zeta }^2}$
$\because \zeta \in (x-\frac{1}{2},x+\frac{1}{2})$
$\therefore \mathrm{当}x\to \mathrm{\infty }\Rightarrow \zeta \to \mathrm{\infty }\mathrm{且}\zeta \sim x$
$\Rightarrow \mathrm{原}\mathrm{式}=\frac{1}{2}\underset{x\to \mathrm{\infty }}{lim}{e}^{\frac{1}{2\zeta }}=\frac{1}{2}.$