这一节引入一个极其重要的求极限方法,洛必达法则。把求极限的内容放到这里,是因为洛必达法则的应用需要导数或微分作为前置知识。
在引入洛必达法则前,先引入柯西中值定理 ,它是证明洛必达法则需要的定理。先给出它的定义:
Suppose functions f and g are continuous on [a, b] and differentiable throughout (a, b) and also suppose g'(x) ≠ \neq = 0 throughout (a, b). Then there exists a number c in (a, b) at which
f ′ ( c ) g ′ ( c ) = f ( b ) − f ( a ) g ( b ) − g ( a ) \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} g ′ ( c ) f ′ ( c ) = g ( b ) − g ( a ) f ( b ) − f ( a )
首先,因为g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 ,那么根据罗尔定理可知,g ( b ) ≠ g ( a ) g(b) \neq g(a) g ( b ) = g ( a ) ,因为如果g ( b ) = g ( a ) g(b) = g(a) g ( b ) = g ( a ) ,那么根据罗尔定理,就会存在至少一个c,使得g'(c) = 0,这和g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 矛盾。所以g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 是保证上面式子的右边分母不为零的关键。
证明:
首先将上式转化为
f ′ ( c ) ( g ( b ) − g ( a ) ) = g ′ ( c ) ( f ( b ) − f ( a ) ) f'(c)(g(b) - g(a)) = g'(c)(f(b) - f(a)) f ′ ( c ) ( g ( b ) − g ( a )) = g ′ ( c ) ( f ( b ) − f ( a ))
将c视为变量x,f(a), f(b), g(a), g(b)均为常数,上式转化为:
f ′ ( x ) ( g ( b ) − g ( a ) ) = g ′ ( x ) ( f ( b ) − f ( a ) ) f'(x)(g(b) - g(a)) = g'(x)(f(b) - f(a)) f ′ ( x ) ( g ( b ) − g ( a )) = g ′ ( x ) ( f ( b ) − f ( a ))
变成要证明:
f ′ ( x ) ( g ( b ) − g ( a ) ) − g ′ ( x ) ( f ( b ) − f ( a ) ) = 0 f'(x)(g(b) - g(a)) - g'(x)(f(b) - f(a)) = 0 f ′ ( x ) ( g ( b ) − g ( a )) − g ′ ( x ) ( f ( b ) − f ( a )) = 0
令:
F ( x ) = f ( x ) ( g ( b ) − g ( a ) ) − g ( x ) ( f ( b ) − f ( a ) ) F(x) = f(x)(g(b) - g(a)) - g(x)(f(b) - f(a)) F ( x ) = f ( x ) ( g ( b ) − g ( a )) − g ( x ) ( f ( b ) − f ( a ))
通过计算可知:
F ( a ) = f ( a ) g ( b ) − g ( a ) f ( b ) F ( b ) = g ( b ) f ( a ) − f ( b ) g ( a ) F ( a ) = F ( b ) F(a) = f(a)g(b) - g(a)f(b) \\
F(b) = g(b)f(a) - f(b)g(a) \\
F(a) = F(b) F ( a ) = f ( a ) g ( b ) − g ( a ) f ( b ) F ( b ) = g ( b ) f ( a ) − f ( b ) g ( a ) F ( a ) = F ( b )
F(x)在a,b区间也是连续可导的,因此应用罗尔定理可知,存在至少一个c,使得F ′ ( c ) = 0 F'(c) = 0 F ′ ( c ) = 0 ,即
F ′ ( c ) = f ′ ( c ) ( g ( b ) − g ( a ) ) − g ′ ( c ) ( f ( b ) − f ( a ) ) = 0 F'(c) = f'(c)(g(b) - g(a)) - g'(c)(f(b) - f(a)) = 0 F ′ ( c ) = f ′ ( c ) ( g ( b ) − g ( a )) − g ′ ( c ) ( f ( b ) − f ( a )) = 0
证毕。
利用拉格朗日中值定理也能证明柯西中值定理。因为f和g有相同的定义域,将其自变量转化为t,那么两函数可以表达为参数函数的形式。x = g ( t ) , y = f ( t ) x = g(t), y = f(t) x = g ( t ) , y = f ( t ) 。根据参数函数的链式求导法则:
d y d x = d y d t d x d t = f ′ ( t ) g ′ ( t ) \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{f'(t)}{g'(t)} d x d y = d t d x d t d y = g ′ ( t ) f ′ ( t )
对新函数y = F(x)应用拉格朗日中值定理,存在至少一个x = g(c),使得:
f ( b ) − f ( a ) g ( b ) − g ( a ) = d y d x ∣ x = g ( c ) = f ′ ( c ) g ′ ( c ) \frac{f(b) - f(a)}{g(b) - g(a)} = \left.\frac{dy}{dx} \right|_{x = g(c)} = \frac{f'(c)}{g'(c)} g ( b ) − g ( a ) f ( b ) − f ( a ) = d x d y x = g ( c ) = g ′ ( c ) f ′ ( c )
证毕。
罗尔定理,拉格朗日中值定理,柯西中值定理都是很重要的定理,它们之间的推导关系可以认为:
朗格朗日中值定理 → \to → 罗尔定理 → \to → 柯西中值定理。
现在开始讨论洛必达法则。
我们求极限的时候,会遇到当趋近某个数值时,表达式变成0/0的情况,它被称为不定式(indeterminate form)。洛必达法则对于这种形式有定理:
Suppose that f(a) = g(a) = 0, that f'(a) and g'(a) exist, and that g'(a) ≠ \neq = 0. Then
lim x → a f ( x ) g ( x ) = f ′ ( a ) g ′ ( a ) \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} x → a lim g ( x ) f ( x ) = g ′ ( a ) f ′ ( a )
这个定理要注意f'(a)和g'(a)要存在,且g ′ ( a ) ≠ 0 g'(a) \neq 0 g ′ ( a ) = 0
证明:
f ′ ( a ) g ′ ( a ) = lim x → a f ( x ) − f ( a ) x − a lim x → a g ( x ) − g ( a ) x − a = lim x → a f ( x ) − f ( a ) x − a g ( x ) − g ( a ) x − a (反过来应用极限除法法则) = lim x → a f ( x ) − f ( a ) g ( x ) − g ( a ) = lim x → a f ( x ) g ( x ) \begin{aligned}
\frac{f'(a)}{g'(a)} &= \frac{\lim_{x \to a}\frac{f(x) - f(a)}{x - a}}{\lim_{x \to a}\frac{g(x) - g(a)}{x - a}} \\
&= \lim_{x \to a}\frac{\frac{f(x) - f(a)}{x - a}}{\frac{g(x) - g(a)}{x - a}}
\ \text{(反过来应用极限除法法则)} \\
&= \lim_{x \to a}\frac{f(x) - f(a)}{g(x) - g(a)} \\
&= \lim_{x \to a}\frac{f(x)}{g(x)}
\end{aligned} g ′ ( a ) f ′ ( a ) = lim x → a x − a g ( x ) − g ( a ) lim x → a x − a f ( x ) − f ( a ) = x → a lim x − a g ( x ) − g ( a ) x − a f ( x ) − f ( a ) ( 反过来应用极限除法法则 ) = x → a lim g ( x ) − g ( a ) f ( x ) − f ( a ) = x → a lim g ( x ) f ( x )
证毕。
但是,有时候f ′ ( a ) f'(a) f ′ ( a ) 和g ′ ( a ) g'(a) g ′ ( a ) 也均为零,那么这时候另一种形式的洛必达法则为:
Suppose that f(a) = g(a) = 0, that f and g are differentiable on an open interval I containing a, and that g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 on I if x ≠ 0 x \neq 0 x = 0 , Then
lim x → a f ( x ) g ( x ) = lim x → a f ′ ( x ) g ′ ( x ) \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} x → a lim g ( x ) f ( x ) = x → a lim g ′ ( x ) f ′ ( x )
assuming that the limit on the right side exists.
洛必达法则有无限“延伸”的特性,因为如果上式成立,在相同条件下可知:
lim x → a f ′ ( x ) g ′ ( x ) = lim x → a f ′ ′ ( x ) g ′ ′ ( x ) \lim_{x \to a}\frac{f'(x)}{g'(x)} = \lim_{x \to a}\frac{f''(x)}{g''(x)} x → a lim g ′ ( x ) f ′ ( x ) = x → a lim g ′′ ( x ) f ′′ ( x )
当然,洛必达法则的大前提是不定式0/0的情况要满足,一旦出现非不定式就不能继续“延伸”了 。
下面来证明下这种形式的洛必达法则:
我们根据极限的定义分开从左极限和右极限来证明。首先假设x → a + x \to a^+ x → a + ,那么x和a之间的区间可以应用柯西中值定理得到:
f ′ ( c ) g ′ ( c ) = f ( x ) − f ( a ) g ( x ) − g ( a ) \frac{f'(c)}{g'(c)} = \frac{f(x) - f(a)}{g(x) - g(a)} g ′ ( c ) f ′ ( c ) = g ( x ) − g ( a ) f ( x ) − f ( a )
由于f(a)和g(a)均为零,上式可以化简为:
f ′ ( c ) g ′ ( c ) = f ( x ) g ( x ) \frac{f'(c)}{g'(c)} = \frac{f(x)}{g(x)} g ′ ( c ) f ′ ( c ) = g ( x ) f ( x )
那么
lim x → a + f ( x ) g ( x ) = lim c → a + f ′ ( c ) g ′ ( c ) \lim_{x \to a^+}\frac{f(x)}{g(x)} = \lim_{c \to a^+}\frac{f'(c)}{g'(c)} x → a + lim g ( x ) f ( x ) = c → a + lim g ′ ( c ) f ′ ( c )
c是一个夹在a和x之间的值,在上式中c是一个变量,因为a和x之间无限趋近,其实c也可以认为等价于x。那么将c替换为x则得到:
lim x → a + f ( x ) g ( x ) = lim x → a + f ′ ( x ) g ′ ( x ) \lim_{x \to a^+}\frac{f(x)}{g(x)} = \lim_{x \to a^+}\frac{f'(x)}{g'(x)} x → a + lim g ( x ) f ( x ) = x → a + lim g ′ ( x ) f ′ ( x )
同理,x → a − x \to a^- x → a − 也可以如上方式证明。
证毕。
从证明方式也可知,洛必达法则也适用于单边极限的求解。
洛必达法则可以做一定的扩展,比如∞ / ∞ \infty/\infty ∞/∞ 的形式也是成立的。又或者对于x → ∞ x \to \infty x → ∞ 也是成立的。
例子:
a. lim x → π / 2 sec x 1 + tan x \lim_{x \to \pi/2}\frac{\sec x}{1 + \tan x} lim x → π /2 1 + t a n x s e c x
lim x → ( π / 2 ) − sec x 1 + tan x , ∞ ∞ = lim x → ( π / 2 ) − sec x tan x sec 2 x = lim x → ( π / 2 ) − sin x = 1 \lim_{x \to (\pi/2)^-}\frac{\sec x}{1 + \tan x}, \ \frac{\infty}{\infty} \\
= \lim_{x \to (\pi/2)^-}\frac{\sec x \tan x}{\sec^2 x} \\
= \lim_{x \to (\pi/2)^-}\sin x = 1 x → ( π /2 ) − lim 1 + tan x sec x , ∞ ∞ = x → ( π /2 ) − lim sec 2 x sec x tan x = x → ( π /2 ) − lim sin x = 1
同样的,右极限也是1.
b. lim x → ∞ x − 2 x 2 3 x 2 + 5 x \lim_{x \to \infty}\frac{x - 2x^2}{3x^2 + 5x} lim x → ∞ 3 x 2 + 5 x x − 2 x 2
lim x → ∞ x − 2 x 2 3 x 2 + 5 x = lim x → ∞ 1 − 4 x 6 x + 5 = lim x → ∞ − 4 6 = − 2 3 \lim_{x \to \infty}\frac{x - 2x^2}{3x^2 + 5x} = \lim_{x \to \infty}\frac{1 - 4x}{6x + 5} = \lim_{x \to \infty}\frac{-4}{6} = \frac{-2}{3} x → ∞ lim 3 x 2 + 5 x x − 2 x 2 = x → ∞ lim 6 x + 5 1 − 4 x = x → ∞ lim 6 − 4 = 3 − 2
有时候还会遇到∞ ⋅ 0 \infty \cdot 0 ∞ ⋅ 0 或者∞ − ∞ \infty - \infty ∞ − ∞ 的形式,这时候就有想办法将其转换为f r a c ∞ ∞ frac{\infty}{\infty} f r a c ∞ ∞ 或0 0 \frac{0}{0} 0 0 的形式。
比如:
a. lim x → ∞ ( x sin 1 x ) \lim_{x \to \infty}(x \sin \frac{1}{x}) lim x → ∞ ( x sin x 1 )
lim x → ∞ ( x sin 1 x ) = lim h → 0 + ( 1 h sin h ) Let h = 1 / x = 1 \lim_{x \to \infty}(x \sin \frac{1}{x}) \\
= \lim_{h \to 0^+}(\frac{1}{h} \sin h) \ \ \text{Let} \ h = 1/x \\
= 1 x → ∞ lim ( x sin x 1 ) = h → 0 + lim ( h 1 sin h ) Let h = 1/ x = 1
b. lim x → 0 ( 1 sin x − 1 x ) \lim_{x \to 0}(\frac{1}{\sin x} - \frac{1}{x}) lim x → 0 ( s i n x 1 − x 1 )
lim x → 0 ( 1 sin x − 1 x ) = lim x → 0 x − sin x x sin x = lim x → 0 1 − cos x sin x + x cos x = lim x → 0 sin x 2 cos x − x sin x = 0 2 = 0 \lim_{x \to 0}(\frac{1}{\sin x} - \frac{1}{x}) \\
= \lim_{x \to 0}\frac{x - \sin x}{x \sin x} \\
= \lim_{x \to 0}\frac{1 - \cos x}{\sin x + x \cos x} \\
= \lim_{x \to 0}\frac{\sin x}{2\cos x - x\sin x} = \frac{0}{2} = 0 x → 0 lim ( sin x 1 − x 1 ) = x → 0 lim x sin x x − sin x = x → 0 lim sin x + x cos x 1 − cos x = x → 0 lim 2 cos x − x sin x sin x = 2 0 = 0
练习
lim x → 2 x − 2 x 2 − 4 \lim_{x \to 2}\frac{x - 2}{x^2 - 4} lim x → 2 x 2 − 4 x − 2
lim x → 2 x − 2 x 2 − 4 = lim x → 2 x − 2 ( x − 2 ) ( x + 2 ) = lim x → 2 1 x + 2 = 1 / 4 \lim_{x \to 2}\frac{x - 2}{x^2 - 4} = \lim_{x \to 2}\frac{x - 2}{(x - 2)(x + 2)} = \lim_{x \to 2}\frac{1}{x + 2} = 1/4 x → 2 lim x 2 − 4 x − 2 = x → 2 lim ( x − 2 ) ( x + 2 ) x − 2 = x → 2 lim x + 2 1 = 1/4
lim x → 2 x − 2 x 2 − 4 = 1 2 ⋅ 2 = 1 / 4 \lim_{x \to 2}\frac{x - 2}{x^2 - 4} = \frac{1}{2 \cdot 2} = 1/4 x → 2 lim x 2 − 4 x − 2 = 2 ⋅ 2 1 = 1/4
lim x → 0 sin 5 x x \lim_{x \to 0}\frac{\sin 5x}{x} lim x → 0 x s i n 5 x
lim x → 0 sin 5 x x = 5 cos ( 5 ⋅ 0 ) 1 = 1 \lim_{x \to 0}\frac{\sin 5x}{x} = \frac{5\cos (5 \cdot 0)}{1} = 1 x → 0 lim x sin 5 x = 1 5 cos ( 5 ⋅ 0 ) = 1
lim x → ∞ 5 x 2 − 3 x 7 x 2 + 1 \lim_{x \to \infty}\frac{5x^2 - 3x}{7x^2 + 1} lim x → ∞ 7 x 2 + 1 5 x 2 − 3 x
lim x → ∞ 5 x 2 − 3 x 7 x 2 + 1 = lim x → ∞ 10 x − 3 14 x = lim x → ∞ 10 14 = 5 \7 \lim_{x \to \infty}\frac{5x^2 - 3x}{7x^2 + 1} = \lim_{x \to \infty}\frac{10x - 3}{14x} = \lim_{x \to \infty}\frac{10}{14} = 5\7 x → ∞ lim 7 x 2 + 1 5 x 2 − 3 x = x → ∞ lim 14 x 10 x − 3 = x → ∞ lim 14 10 = 5 \7
lim x → 1 x 3 − 1 4 x 3 − x − 3 \lim_{x \to 1}\frac{x^3 - 1}{4x^3 - x - 3} lim x → 1 4 x 3 − x − 3 x 3 − 1
lim x → 1 x 3 − 1 4 x 3 − x − 3 = lim x → 1 3 x 2 12 x 2 − 1 = 3 / 11 \lim_{x \to 1}\frac{x^3 - 1}{4x^3 - x - 3} = \lim_{x \to 1}\frac{3x^2}{12x^2 - 1} = 3/11 x → 1 lim 4 x 3 − x − 3 x 3 − 1 = x → 1 lim 12 x 2 − 1 3 x 2 = 3/11
lim x → 0 1 − cos x x 2 \lim_{x \to 0}\frac{1 - \cos x}{x^2} lim x → 0 x 2 1 − c o s x
lim x → 0 1 − cos x x 2 = lim x → 0 sin x 2 x = lim x → 0 − cos x 2 = − 1 / 2 \lim_{x \to 0}\frac{1 - \cos x}{x^2} = \lim_{x \to 0}\frac{\sin x}{2x} = \lim_{x \to 0}\frac{-\cos x}{2} = -1/2 x → 0 lim x 2 1 − cos x = x → 0 lim 2 x sin x = x → 0 lim 2 − cos x = − 1/2
lim x → ∞ 2 x 2 + 3 x x 3 + x + 1 \lim_{x \to \infty}\frac{2x^2 + 3x}{x^3 + x + 1} lim x → ∞ x 3 + x + 1 2 x 2 + 3 x
lim x → ∞ 2 x 2 + 3 x x 3 + x + 1 = lim x → ∞ 4 x + 3 3 x 2 + 1 = lim x → ∞ 4 6 x = 0 \lim_{x \to \infty}\frac{2x^2 + 3x}{x^3 + x + 1} = \lim_{x \to \infty}\frac{4x + 3}{3x^2 + 1} = \lim_{x \to \infty}\frac{4}{6x} = 0 x → ∞ lim x 3 + x + 1 2 x 2 + 3 x = x → ∞ lim 3 x 2 + 1 4 x + 3 = x → ∞ lim 6 x 4 = 0