【数学】有理分式求极限

一开始就略过了这个部分,然后导致后面求分式极限都十分不好操作,现在补回来。
设多项式
\[f(x)=\sum_{i=0}^{n}a_{i}x^{n-i} \]
则:
\[\lim_{x \to x_{0}}f(x)=\lim_{x \to x_{0}}(\sum_{i=0}^{n}a_{i}x^{n-i})\]
\[=\sum_{i=0}^{n}a_{i}\left(\lim_{x \to x_{0}}x \right)^{n-i}\]
\[\sum_{i=0}^{n}a_{i}x_{0}^{n-i}=f(x_{0})\]
又设有理分式函数
\[F(x)=\frac{P(x)}{Q(x)}\]
其中$P(x),Q(x)$都是多项式,于是:
\[\lim_{x \to x_{0}}P(x)=P(x_{0}),\lim_{x \to x_{0}}Q(x)=Q(x_{0})\]
如果$Q(x_{0})\neq 0$,那么:
\[\lim_{x \to x_{0}}F(x)=\lim_{x \to x_{0}}\frac{P(x)}{Q(x)}=\frac{\lim_{x \to x_{0}}P(x)}{\lim_{x \to x_{0}}Q(x)}=F(x_{0})\]