常用极限
| lim x → 0 sin x x = 1 \lim_{x \to 0} {\frac{\sin x}{x}}=1 limx→0xsinx=1 | lim x → 0 ( x + 1 ) 1 x = e \lim_{x \to 0} {(x+1)^\frac{1}{x}}=e limx→0(x+1)x1=e | lim n → ∞ a n = 1 \lim_{n \to \infty} {\sqrt[n]{a}}=1 limn→∞na=1 |
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| lim n → ∞ n n = 1 \lim_{n \to \infty} {\sqrt[n]{n}}=1 limn→∞nn=1 | lim x → + ∞ arctan x = π 2 \lim_{x \to +\infty} {\arctan x}=\frac{\pi}{2} limx→+∞arctanx=2π | lim x → − ∞ arctan x = − π 2 \lim_{x \to -\infty} {\arctan x}=-\frac{\pi}{2} limx→−∞arctanx=−2π |
| lim x → + ∞ arccot x = 0 \lim_{x \to +\infty} {\text{arccot} x}=0 limx→+∞arccotx=0 | lim x → − ∞ arccot x = π \lim_{x \to -\infty} {\text{arccot} x}=\pi limx→−∞arccotx=π | lim x → − ∞ e x = 0 \lim_{x \to -\infty} {e^x}=0 limx→−∞ex=0 |
| lim x → + ∞ e x = ∞ \lim_{x \to +\infty} {e^x}=\infty limx→+∞ex=∞ | lim x → 0 + x x = 1 \lim_{x \to 0^+} {x^x}=1 limx→0+xx=1 | lim x → ∞ ( 1 x + 1 ) x = e \lim_{x \to \infty} {(\frac{1}{x}+1)^x}=e limx→∞(x1+1)x=e |
常用等价无穷小： x → 0 x\to 0 x→0，替换条件：只能分子及分母中因式分解可替换
| sin x ∼ x \sin x \sim x sinx∼x | tan x ∼ x \tan x \sim x tanx∼x | arcsin x ∼ x \arcsin x \sim x arcsinx∼x |
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| arctan x ∼ x \arctan x \sim x arctanx∼x | 1 − cos x ∼ 1 2 x 2 1-\cos x \sim \frac{1}{2}x^2 1−cosx∼21x2 | ln ( 1 + x ) ∼ x \ln(1+x) \sim x ln(1+x)∼x |
| e x − 1 ∼ x e^x-1 \sim x ex−1∼x | a x − 1 ∼ x ln a a^x-1 \sim x\ln a ax−1∼xlna | ( 1 + x ) α ∼ α x (1+x)^\alpha \sim \alpha x (1+x)α∼αx |
| sin x − x ∼ − 1 6 x 3 \sin x -x \sim -\frac{1}{6}x^3 sinx−x∼−61x3 |
等价无穷小替换注意：
1、被替换的量必须是无穷小量（取极限为0）
例： lim x → 0 x 2 ∗ sin 1 x 2 \lim_{x \to 0} x^2*\sin \frac{1}{x^2} limx→0x2∗sinx21
∵ lim x → 0 x 2 = 0 ， lim x → 0 sin 1 x 2 极限不存在且为有界函数 \because \quad \lim_{x \to 0} x^2=0， \lim_{x \to 0} \sin \frac{1}{x^2} 极限不存在且为有界函数 ∵limx→0x2=0，limx→0sinx21极限不存在且为有界函数
∴ 无穷小与有界的集为无穷小， lim x → 0 x 2 ∗ sin 1 x 2 = 0 \therefore \quad 无穷小与有界的集为无穷小，\lim_{x \to 0} x^2*\sin \frac{1}{x^2}=0 ∴无穷小与有界的集为无穷小，limx→0x2∗sinx21=0
2、被替换的量，必须是被乘式被除的元素
例： lim x → 0 sin x − tan x x 3 \lim_{x \to 0} \frac{\sin x - \tan x}{x^3} limx→0x3sinx−tanx
∵ lim x → 0 sin x − tan x x 3 = lim x → 0 tan x ( cos x − 1 ) x 3 ， lim x → 0 tan x x = 1 , 1 − cos x ∼ 1 2 x 2 \because \lim_{x \to 0} \frac{\sin x - \tan x}{x^3} = \lim_{x \to 0} \frac{\tan x(\cos x - 1)}{x^3}，\lim_{x \to 0} {\frac{\tan x}{x}}=1,1-\cos x \sim \frac{1}{2}x^2 ∵limx→0x3sinx−tanx=limx→0x3tanx(cosx−1)，limx→0xtanx=1,1−cosx∼21x2
∴ lim x → 0 sin x − tan x x 3 = − 1 ∗ 1 2 x 2 x 2 = − 1 2 \therefore \quad \lim_{x \to 0} \frac{\sin x - \tan x}{x^3} =-1* \frac{\frac{1}{2}x^2}{x^2}=-\frac{1}{2} ∴limx→0x3sinx−tanx=−1∗x221x2=−21
3、必须整体替换：不能替换局部
例： lim x → 0 sin x ( x + 1 ) x \lim_{x \to 0} \frac{\sin x(x+1)}{x} limx→0xsinx(x+1)
∵ lim x → 0 sin x ( x + 1 ) x = lim x → 0 1 ∗ ( 1 + x ) = 0 \because \quad \lim_{x \to 0} \frac{\sin x(x+1)}{x}= \lim_{x \to 0} 1*(1+x)=0 ∵limx→0xsinx(x+1)=limx→01∗(1+x)=0
例： lim x → 0 ( e x + x e x e x − 1 − 1 x ) \lim_{x \to 0} (\frac{e^x+xe^x}{e^x-1} - \frac{1}{x}) limx→0(ex−1ex+xex−x1)
∵ lim x → 0 ( e x + x e x e x − 1 − 1 x ) = lim x → 0 ( e x ( 1 + x ) e x − 1 − 1 x ) ∗ lim x → 0 e x − 1 x = lim x → 0 e x ( x 2 + x − 1 ) + 1 x 2 \because \quad \lim_{x \to 0} (\frac{e^x+xe^x}{e^x-1} - \frac{1}{x})=\lim_{x \to 0}(\frac{e^x(1+x)}{e^x-1} - \frac{1}{x})* \lim_{x \to 0} \frac{e^x-1}{x}=\lim_{x \to 0}\frac{e^x(x^2+x-1)+1}{x^2} ∵limx→0(ex−1ex+xex−x1)=limx→0(ex−1ex(1+x)−x1)∗limx→0xex−1=limx→0x2ex(x2+x−1)+1
∴ 上述是 0 0 , 对 lim x → 0 e x ( x 2 + x − 1 ) + 1 x 2 上下同时求导得 \therefore \quad 上述是\frac{0}{0},对\lim_{x \to 0} \frac{e^x(x^2+x-1)+1}{x^2}上下同时求导得 ∴上述是00,对limx→0x2ex(x2+x−1)+1上下同时求导得
lim x → 0 e x ( x 2 + x − 1 ) + 1 x 2 = lim x → 0 e x ( x 2 + 3 x ) 2 x = 3 2 \quad \quad \lim_{x \to 0} \frac{e^x(x^2+x-1)+1}{x^2}=\lim_{x \to 0} \frac{e^x(x^2+3x)}{2x}=\frac{3}{2} limx→0x2ex(x2+x−1)+1=limx→02xex(x2+3x)=23
