洛必达法则的一种极简证明

洛必达法则具体如下： 如有： lim ⁡ x → a f ( x ) = 0 , lim ⁡ x → a g ( x ) = 0      o r      lim ⁡ x → a f ( x ) = ∞ , lim ⁡ x → a g ( x ) = ∞ \lim _{x \rightarrow a} f(x)=0, \lim _{x \rightarrow a} g(x)=0 \;\;\mathrm{or}\;\; \lim _{x \rightarrow a} f(x)=\infty, \lim _{x \rightarrow a} g(x)=\infty x→alim​f(x)=0,x→alim​g(x)=0orx→alim​f(x)=∞,x→alim​g(x)=∞ 则有（默认邻域内可导： lim ⁡ x → a f ( x ) g ( x ) = lim ⁡ x → a f ′ ( x ) g ′ ( x ) = A \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=A x→alim​g(x)f(x)​=x→alim​g′(x)f′(x)​=A

证明： 先考虑 lim ⁡ x → a f ( x ) = 0 , lim ⁡ x → a g ( x ) = 0 \lim _{x \rightarrow a} f(x)=0, \lim _{x \rightarrow a} g(x)=0 limx→a​f(x)=0,limx→a​g(x)=0. 即： f ( a ) = g ( a ) = 0 f(a)=g(a)=0 f(a)=g(a)=0 根据柯西中值定理，存在 m m m为 ( a , x ) (a,x) (a,x)间一点， 有： f ( x ) g ( x ) = f ( x ) − f ( a ) g ( x ) − g ( a ) = f ′ ( m ) g ′ ( m ) \frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^\prime(m)}{g^\prime (m)} g(x)f(x)​=g(x)−g(a)f(x)−f(a)​=g′(m)f′(m)​ 令 x → a x\rightarrow a x→a， 此时 m → a m\rightarrow a m→a， 得到： lim ⁡ x → a f ( x ) g ( x ) = lim ⁡ x → a f ′ ( x ) g ′ ( x ) . \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a}\frac{f^\prime(x)}{g^\prime (x)}. x→alim​g(x)f(x)​=x→alim​g′(x)f′(x)​.

接下来考虑 lim ⁡ x → a f ( x ) = ∞ , lim ⁡ x → a g ( x ) = ∞ \lim _{x \rightarrow a} f(x)=\infty, \lim _{x \rightarrow a} g(x)=\infty limx→a​f(x)=∞,limx→a​g(x)=∞， 我们首先有： f ( x ) g ( x ) = 1 g ( x ) 1 f ( x ) \frac{f(x)}{g(x)}=\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}} g(x)f(x)​=f(x)1​g(x)1​​, 显然当 x → a x \rightarrow a x→a时， 分子分母都趋近于0， 因此根据刚证明的结论，有：

lim ⁡ x → a 1 g ( x ) 1 f ( x ) = lim ⁡ x → a ( 1 g ( x ) ) ′ ( 1 f ( x ) ) ′ = lim ⁡ x → a − 1 g ( x ) 2 g ( x ) ′ − 1 f ( x ) 2 f ( x ) ′ = lim ⁡ x → a 1 g 2 ( x ) 1 f 2 ( x ) lim ⁡ x → a g ( x ) ′ f ( x ) ′ \lim _{x \rightarrow a}\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}=\lim _{x \rightarrow a}\frac{(\frac{1}{g(x)})^\prime}{(\frac{1}{f(x)})^\prime}=\lim _{x \rightarrow a}\frac{-\frac{1}{g(x)^2}g(x)^\prime}{-\frac{1}{f(x)^2}f(x)^\prime}=\lim _{x \rightarrow a}\frac{\frac{1}{g^2(x)}}{\frac{1}{f^2(x)}}\lim _{x \rightarrow a}\frac{g(x)^\prime}{f(x)^\prime} x→alim​f(x)1​g(x)1​​=x→alim​(f(x)1​)′(g(x)1​)′​=x→alim​−f(x)21​f(x)′−g(x)21​g(x)′​=x→alim​f2(x)1​g2(x)1​​x→alim​f(x)′g(x)′​, 两边都除以 t 2 t^2 t2, t = lim ⁡ x → a 1 g ( x ) 1 f ( x ) t=\lim _{x \rightarrow a}\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}} t=limx→a​f(x)1​g(x)1​​, 有： lim ⁡ x → a g ( x ) f ( x ) = lim ⁡ x → a g ( x ) ′ f ( x ) ′ \lim_{x \rightarrow a}\frac{g(x)}{f(x)}=\lim _{x \rightarrow a}\frac{g(x)^\prime}{f(x)^\prime} x→alim​f(x)g(x)​=x→alim​f(x)′g(x)′​

证毕。