# 极限

Python中求该极限方法如下：

#### 泰勒级数用于极限计算

$lim_{xrightarrow 0}frac{sin(x)}{x}=lim_{xrightarrow 0}{frac{frac{x}{1!}-frac{x^3}{3!}+frac{x^5}{5!}-frac{x^7}{7!}+dots}{x}}$
$qquad = lim_{xrightarrow 0}{frac{x(1-frac{x^2}{3!}+frac{x^4}{5!}-frac{x^6}{7!}+dots)}{x}}$
$qquad = lim_{xrightarrow 0}{1-frac{x^2}{3!}+frac{x^4}{5!}-frac{x^6}{7!}+dots}$
$qquad = 1$

#### 洛必达法则（l'Hopital's rule）

$lim_{xrightarrow a}frac{f(x)}{g(x)}=lim_{xrightarrow a}frac{f'(x)}{g'(x)}$

$lim_{xrightarrow a}{frac{f(x)}{g(x)}}=lim_{xrightarrow a}{frac{f(a)+frac{f'(a)}{1!}(x-a)+frac{f''(a)}{2!}(x-a)^2+frac{f'''(a)}{3!}(x-a)^3+dots}{g(a)+frac{g'(a)}{1!}(x-a)+frac{g''(a)}{2!}(x-a)^2+frac{g'''(a)}{3!}(x-a)^3+dots}}$
$qquad = lim_{xrightarrow a}{frac{frac{f'(a)}{1!}(x-a)+frac{f''(a)}{2!}(x-a)^2+frac{f'''(a)}{3!}(x-a)^3+dots}{frac{g'(a)}{1!}(x-a)+frac{g''(a)}{2!}(x-a)^2+frac{g'''(a)}{3!}(x-a)^3+dots}}$
$qquad =lim_{xrightarrow a}{frac{f'(a)+frac{f''(a)}{2!}(x-a)+frac{f'''(a)}{3!}(x-a)^2+dots}{g'(a)+frac{g''(a)}{2!}(x-a)+frac{g'''(a)}{3!}(x-a)^2+dots}}$
$qquad = lim_{xrightarrow a}frac{f'(x)}{g'(x)}$