重要极限及其推论

三角函数 $\frac{0}{0}$ 型

$\lim_{x \to 0} \dfrac{\sin x}{x} = 1$, ($\lim_{x \to \infty} \dfrac{\sin x}{x} = 0$)

推论:

1. $\lim_{x \to x_0} \dfrac{\sin \varphi(x_0)}{\varphi(x_0)} = 1, (\lim_{x \to x_0} \varphi(x)=0, \varphi(x) \neq 0)$
2. $\lim_{x \to 0} \dfrac{\tan x}{x} = \lim_{x \to 0} \dfrac{\sin x}{x\cos x} = 1$
3. $\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \lim_{x \to 0} \dfrac{2\sin^2 \dfrac{x}{2}}{(\dfrac{x}{2})^2 \cdot 4} = \dfrac{1}{2}$
4. $\lim_{x \to 0} \dfrac{\arcsin x}{x} = \lim_{x \to 0} \dfrac{\arcsin x}{\sin(\arcsin x)} = 1$
5. $\lim_{x \to 0} \dfrac{\arctan x}{x} = \lim_{x \to 0} \dfrac{\arctan x}{\tan(\arctan x)} = 1$

对数/指数/幂 $1^\infty$ 型

$\lim_{x \to \infty} (1 + \dfrac{1}{x})^x = e, \lim_{x \to 0} (1 + x)^\dfrac{1}{x} = e$$\implies \lim_{x \to \infty} (1 - \dfrac{1}{x})^x = \lim_{x \to \infty} (1 + \dfrac{1}{x})^{-x} = \dfrac{1}{e}$

推论:

1. $\lim_{x \to x_0} (1 + \varphi(x))^\dfrac{1}{\varphi(x)} = e$, $(\lim_{x \to x_0} \varphi(x) = \infty)$
2. $\lim_{x \to 0} \dfrac{\ln (1+x)}{x} = \lim_{x \to 0} \ln (1 + x)^\dfrac{1}{x} = \ln e = 1$
3. $\lim_{x \to 0} \dfrac{e^x-1}{x} = \lim_{t \to 0} \dfrac{t}{\ln (1+t)} = 1$, $(t=e^x - 1, x=\ln(t+1))$ 换元法 ($x \to 0$ 时 $t \to 0$)
4. $\lim_{x \to 0} \dfrac{a^x - 1}{x} = \lim_{t \to 0} \dfrac{t}{\ln(1+t)}\ln a = \ln a$, $(t=a^x-1, x=\dfrac{\ln(1+t)}{\ln a})$ 换元法 ($x \to 0$ 时 $t \to 0$)
5. $\lim_{x \to 0} \dfrac{(1+x)^\alpha - 1}{x} = \lim_{x \to 0, t \to 0} \dfrac{t}{\ln (1+t)} \cdot \dfrac{\alpha \ln(1+x)}{x} = \alpha$, $(t=(1+x)^\alpha - 1, \ln(1+t)=\alpha \ln(1+x))$ 换元法 ($x \to 0$ 时 $t \to 0$)

应用：幂函数 $f(x)^{g(x)} = e^{g(x) \ln f(x)} = e^{g(x) \frac{\ln(1+f(x)-1)}{f(x)-1} [f(x)-1]} \to e^{g(x)[f(x)-1]}$, $(f(x) \to 1)$$\lim f(x)^{g(x)} = e^{\lim g(x) \ln f(x)} = [\lim f(x)]^{\lim g(x)}$, $(f(x) \to A>0)$

警告

注意: $f(x), g(x)$ 必须同时取极限, $\lim f(x)^{g(x)} \neq [\lim f(x)]^{g(x)}$ (特别是 $\lim f(x) = e$ 时易错)