# 洛必达法则

## 敘述

${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$

（1）${\displaystyle \lim _{x\to c}f(x)g(x)\!}$ ${\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {f(x)}{1/g(x)}}\!}$
（2）${\displaystyle \lim _{x\to c}(f(x)-g(x))\!}$ ${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}(f(x)-g(x))=\lim _{x\to c}{\frac {1/g(x)-1/f(x)}{1/(f(x)g(x))}}\!}$
（3）${\displaystyle \lim _{x\to c}{f(x)}^{g(x)}\!}$ ${\displaystyle \lim _{x\to c}f(x)=0^{+},\lim _{x\to c}g(x)=0\!}$
${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=0\!}$
${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!}$
（4）${\displaystyle \lim _{x\to c}{f(x)}^{g(x)}\!}$ ${\displaystyle \lim _{x\to c}f(x)=1,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!}$

## 證明

${\displaystyle f(a)=0;\;g(a)=0,\qquad \lim _{x\to a}f(x)=0;\;\lim _{x\to a}g(x)=0}$

${\displaystyle \lim _{x\to a}{\frac {f'(x)}{g'(x)}}=L.}$

${\displaystyle L-\epsilon \leqslant {\frac {f'(x)}{g'(x)}}\leqslant L+\epsilon }$

 ${\displaystyle {\frac {f(x)}{g(x)}}}$ ${\displaystyle ={\frac {f(x)-f(a)}{g(x)-g(a)}}={\frac {f'(\xi )}{g'(\xi )}}}$ 于是， ${\displaystyle L-\epsilon \leqslant {\frac {f(x)}{g(x)}}\leqslant L+\epsilon }$

## 例子

 ${\displaystyle \lim _{x\to 0}\mathrm {sinc} (x)\,}$ ${\displaystyle =\lim _{x\to 0}{\frac {\sin \pi x}{\pi x}}\,}$ ${\displaystyle =\lim _{x\to 0}{\frac {\sin x}{x}}\,}$ ${\displaystyle =\lim _{x\to 0}{\frac {\cos x}{1}}\,}$ ${\displaystyle ={\frac {1}{1}}=1\,}$
 ${\displaystyle \lim _{x\to 0}{2\sin x-\sin 2x \over x-\sin x}}$ ${\displaystyle =\lim _{x\to 0}{2\cos x-2\cos 2x \over 1-\cos x}}$ ${\displaystyle =\lim _{x\to 0}{-2\sin x+4\sin 2x \over \sin x}}$ ${\displaystyle =\lim _{x\to 0}{-2\cos x+8\cos 2x \over \cos x}}$ ${\displaystyle ={-2\cos 0+8\cos 0 \over \cos 0}}$ ${\displaystyle =6\,}$
 ${\displaystyle \lim _{x\to 0}{r^{x}-1 \over x}}$ ${\displaystyle =\lim _{x\to 0}{{\frac {d}{dx}}r^{x} \over {\frac {d}{dx}}x}}$ ${\displaystyle =\lim _{x\to 0}{r^{x}\ln r \over 1}}$ ${\displaystyle =\ln r\lim _{x\to 0}{r^{x}}}$ ${\displaystyle =\ln r\!}$
${\displaystyle \lim _{x\to 0}{e^{x}-1-x \over x^{2}}=\lim _{x\to 0}{e^{x}-1 \over 2x}=\lim _{x\to 0}{e^{x} \over 2}={1 \over 2}}$
${\displaystyle \lim _{x\to \infty }{\frac {\sqrt {x}}{\ln(x)}}=\lim _{x\to \infty }{\frac {\ {\frac {1}{2{\sqrt {x}}}}\ }{\frac {1}{x}}}=\lim _{x\to \infty }{\frac {\sqrt {x}}{2}}=\infty }$
${\displaystyle \lim _{x\to \infty }x^{n}e^{-x}=\lim _{x\to \infty }{x^{n} \over e^{x}}=\lim _{x\to \infty }{nx^{n-1} \over e^{x}}=n\lim _{x\to \infty }{x^{n-1} \over e^{x}}=0}$
${\displaystyle \lim _{x\to 0+}(x\ln x)=\lim _{x\to 0+}{\ln x \over {\frac {1}{x}}}=\lim _{x\to 0+}{{\frac {1}{x}} \over -{\frac {1}{x^{2}}}}=\lim _{x\to 0+}-x=0}$
 ${\displaystyle \lim _{t\to 0}\,\mathrm {sinc} (f_{0}t)\cdot {\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}}$ ${\displaystyle =\left\{\lim _{t\to 0}\,\mathrm {sinc} (f_{0}t)\right\}\cdot \left.{\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}\,\right|_{t=0}}$ ${\displaystyle =1\cdot 1=1}$
 ${\displaystyle \lim _{t\to {\frac {1}{2\alpha f_{0}}}}\mathrm {sinc} (f_{0}t)\cdot {\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}}$ ${\displaystyle =\mathrm {sinc} \left({\frac {1}{2\alpha }}\right)\cdot \lim _{t\to {\frac {1}{2\alpha f_{0}}}}{\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}}$ ${\displaystyle =\mathrm {sinc} \left({\frac {1}{2\alpha }}\right)\cdot \left({\frac {\frac {-\pi }{2}}{-2}}\right)}$ ${\displaystyle =\sin \left({\frac {\pi }{2\alpha }}\right)\cdot {\frac {\alpha }{2}}}$