# 洛必达法则

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

## 叙述

### 第一种形式

1. $x\to a$时，函数f(x)与F(x)都趋于零；
2. 在点a的某去心邻域内，$f^\prime(x)$$F^\prime(x)$都存在，且$F^\prime(x)\ne 0$
3. $\lim_{x\to a}{\frac{f^\prime(x)}{F^\prime(x)}}$存在（或為無窮大），

$\lim_{x\to a}{\frac{f(x)}{F(x)}}=\lim_{x\to a}{\frac{f^\prime(x)}{F^\prime(x)}}$

### 第二种形式

1. $x\to\infty$时，函数f(x)与F(x)都趋于零；
2. $\left|x\right|>N$时，$f^\prime(x)$$F^\prime(x)$都存在，且$F^\prime(x)\ne 0$
3. $\lim_{x\to \infty}{\frac{f^\prime(x)}{F^\prime(x)}}$存在（或为无穷大），

$\lim_{x\to \infty}{\frac{f(x)}{F(x)}}=\lim_{x\to \infty}{\frac{f^\prime(x)}{F^\prime(x)}}.$

$\frac{\infty}{\infty} = \frac{1/\infty}{1/\infty} = \frac00 .$

$0 \cdot \infty$型不定式极限，可以转化如下：

$0 \cdot \infty = \frac{0}{1/\infty} = \frac00 .$

$\infty - \infty$型不定式极限，可以转化如下：

$\infty - \infty^\prime = \frac{1}{0} - \frac{1}{0^\prime} = \frac{0^\prime - 0}{0 \cdot 0^\prime} .$

$0^0$型不定式极限，可以转化如下：

$0^0 = 0\div0 = \frac{0}{0} .$

## 證明

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

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

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

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

## 例子

 $\lim_{x \to 0} \mathrm{sinc}(x)\,$ $= \lim_{x \to 0} \frac{\sin \pi x}{\pi x}\,$ $= \lim_{x \to 0} \frac{\sin x}{x}\,$ $= \lim_{x \to 0} \frac{\cos x}{1} \,$ $= \frac{1}{1} = 1\,$
 $\lim_{x\to 0} {2\sin x-\sin 2x \over x-\sin x}$ $=\lim_{x\to 0}{2\cos x-2\cos 2x \over 1-\cos x}$ $=\lim_{x\to 0}{-2\sin x +4\sin 2x \over \sin x}$ $=\lim_{x\to 0}{-2\cos x +8\cos 2x \over \cos x}$ $={-2\cos 0 +8\cos 0 \over \cos 0}$ $=6\,$
 $\lim_{x\to 0} {r^x - 1 \over x}$ $=\lim_{x \to 0}{\frac{d}{dx}r^x \over \frac{d}{dx}x}$ $=\lim_{x \to 0}{r^x \ln r \over 1}$ $=\ln r \lim_{x \to 0}{r^x}$ $=\ln r\!$
$\lim_{x\to 0}{e^x-1-x \over x^2} =\lim_{x\to 0}{e^x-1 \over 2x} =\lim_{x\to 0}{e^x \over 2}={1 \over 2}$
$\lim_{x \to \infty} \frac{\sqrt{x}}{\ln(x)} = \lim_{x \to \infty} \frac{\ 1/(2 \sqrt{x})\ }{1/x} = \lim_{x \to \infty} \frac{\sqrt{x}}{2} = \infty$
$\lim_{x\to\infty} x^n e^{-x} =\lim_{x\to\infty}{x^n \over e^x} =\lim_{x\to\infty}{nx^{n-1} \over e^x} =n\lim_{x\to\infty}{x^{n-1} \over e^x}= 0$
$\lim_{x\to 0+} (x \ln x) =\lim_{x\to 0+}{\ln x \over 1/x} =\lim_{x\to 0+}{1/x \over -1/x^2} =\lim_{x\to 0+} -x = 0$
 $\lim_{t\to 0}\, \mathrm{sinc}(f_0 t)\cdot \frac{\cos\left(\pi \alpha f_0 t\right)}{\left[1 - \left(2 \alpha f_0 t\right)^2\right]}$ $= \left\{\lim_{t\to 0}\, \mathrm{sinc}(f_0 t)\right\}\cdot \left. \frac{\cos\left(\pi \alpha f_0 t\right)}{\left[1 - \left(2 \alpha f_0 t\right)^2\right]} \, \right|_{t = 0}$ $= 1 \cdot 1 = 1$
 $\lim_{t\to \frac{1}{2\alpha f_0}} \mathrm{sinc}(f_0 t)\cdot \frac{\cos\left(\pi \alpha f_0 t\right)}{\left[1 - \left(2 \alpha f_0 t\right)^2\right]}$ $= \mathrm{sinc}\left(\frac{1}{2\alpha}\right)\cdot \lim_{t\to \frac{1}{2\alpha f_0}} \frac{\cos\left(\pi \alpha f_0 t\right)}{\left[1 - \left(2 \alpha f_0 t\right)^2\right]}$ $= \mathrm{sinc}\left(\frac{1}{2\alpha}\right)\cdot \left(\frac{-\pi /2}{-2}\right)$ $= \sin\left(\frac{\pi}{2\alpha}\right)\cdot \frac{\alpha}{2}$