0/0型洛必达法则

在区间(a, b)上,f(x)和g(x)都可导、g(x) ≠ 0、limx → a+f(x) = limx → a+g(x) = 0,

$$lim_{x ightarrow a^{+}}frac{fleft( x ight)}{gleft( x ight)} = lim_{x ightarrow a^{+}}frac{f^{'}left( x ight)}{g^{'}left( x ight)}  $$

证明:设f(a)=g(a)=0,则有limx → a+f(x) = f(a) = 0、limx → a+g(x) = g(a) = 0,所以这个定义使得f(x)和g(x)在[a, b)上连续。取任意的x ∈ (ab),由于f(x)和g(x)在[a, x]上满足使用柯西中值定理的条件,所以有

$$frac{f(x) - f(a)}{gleft( x ight) - gleft( a ight)} = frac{f'(c)}{g^{'}left( c ight)}$$

因为f(a)=g(a)=0,所以

$$frac{f(x)}{gleft( x ight)} = frac{f'(c)}{g^{'}left( c ight)}$$

x → a+时,因为c在(a, x)上,所以c → a+,所以

$$lim_{x ightarrow a^{+}}frac{fleft( x ight)}{gleft( x ight)} = lim_{x ightarrow a^{+}}frac{f^{'}left( c ight)}{g^{'}left( c ight)} = lim_{c ightarrow a^{+}}frac{f^{'}left( c ight)}{g^{'}left( c ight)} = lim_{x ightarrow a^{+}}frac{f^{'}left( x ight)}{g^{'}left( x ight)}  $$

原文地址:https://www.cnblogs.com/valar-morghulis/p/13522657.html