两个重要极限： ① lim ⁡ x → 0 sin ⁡ x x = 1 \lim_{x \to 0}\frac{\sin x}{x} = 1 x→0lim​xsinx​=1 ② lim ⁡ x → ∞ ( 1 + 1 x ) x = e \lim_{x \to \infty}(1 + \frac{1}{x})^x = e x→∞lim​(1+x1​)x=e

关于重要极限①的推导极限可以参考： 无穷小的等价代换

关于重要极限②的由来可以参考：自然常数e与重要极限

由重要极限②可以推导出： lim ⁡ x → ∞ ( 1 + 1 x ) x ⇒ lim ⁡ x → 0 ( 1 + x ) 1 x = e \lim_{x \to \infty}(1 + \frac{1}{x})^x \Rightarrow \lim_{x \to 0}(1 + x)^{\frac{1}{x}} = e x→∞lim​(1+x1​)x⇒x→0lim​(1+x)x1​=e

重要极限②的例题 例：求 lim ⁡ x → ∞ ( 1 + 2 x + 3 ) x \lim_{x \to \infty} (1 + \frac{2}{x + 3})^x limx→∞​(1+x+32​)x 解： lim ⁡ x → ∞ ( 1 + 2 x + 3 ) x \lim_{x \to \infty} (1 + \frac{2}{x + 3})^x x→∞lim​(1+x+32​)x = lim ⁡ x → ∞ ( 1 + 1 x + 3 2 ) ( x + 3 2 ) ⋅ 2 − 3 = \lim_{x \to \infty} (1 + \frac{1}{\frac{x + 3}{2}})^{(\frac{x+3}{2})·2 - 3} =x→∞lim​(1+2x+3​1​)(2x+3​)⋅2−3 = lim ⁡ x → ∞ ( 1 + 1 x + 3 2 ) x + 3 2 ⋅ 2 ⋅ ( 1 + 1 x + 3 2 ) − 3 = \lim_{x \to \infty} (1 + \frac{1}{\frac{x + 3}{2}})^{\frac{x+3}{2}·2} ·(1 + \frac{1}{\frac{x + 3}{2}})^{-3} =x→∞lim​(1+2x+3​1​)2x+3​⋅2⋅(1+2x+3​1​)−3 = lim ⁡ x → ∞ ( 1 + 1 x + 3 2 ) x + 3 2 ⋅ 2 =\lim_{x \to \infty} (1 + \frac{1}{\frac{x + 3}{2}})^{\frac{x+3}{2}·2} =x→∞lim​(1+2x+3​1​)2x+3​⋅2 = e 2 = e^2 =e2

关于重要极限②，还有一个变体例题，很容易迷惑人，但实际上并不是用重要极限的方法来做 例：求 lim ⁡ x → ∞ ( 1 + 2 x ) 1 x \lim_{x \to \infty}(1+2x)^{\frac{1}{x}} x→∞lim​(1+2x)x1​ 解： ( 1 + 2 x ) 1 x = e ln ⁡ ( 1 + 2 x ) 1 x = e ln ⁡ ( 1 + 2 x ) x (1+2x)^{\frac{1}{x}} = e^{\ln(1+2x)^{\frac{1}{x}}} = e^{\frac{\ln (1+2x)}{x}} (1+2x)x1​=eln(1+2x)x1​=exln(1+2x)​ 所以： lim ⁡ x → ∞ ( 1 + 2 x ) 1 x = lim ⁡ x → ∞ e ln ⁡ ( 1 + 2 x ) x = e lim ⁡ x → ∞ ln ⁡ ( 1 + 2 x ) x \lim_{x \to \infty}(1+2x)^{\frac{1}{x}} = \lim_{x \to \infty}e^{\frac{\ln (1+2x)}{x}} = e^{\lim_{x \to \infty} \frac{\ln (1+2x)}{x}} x→∞lim​(1+2x)x1​=x→∞lim​exln(1+2x)​=elimx→∞​xln(1+2x)​ 应用洛必达法则，得： e lim ⁡ x → ∞ ln ⁡ ( 1 + 2 x ) x = e lim ⁡ x → ∞ 2 1 + 2 x = e 0 = 1 e^{\lim_{x \to \infty} \frac{\ln (1+2x)}{x}} = e^{\lim_{x \to \infty} \frac{2}{1+2x}} = e^0 = 1 elimx→∞​xln(1+2x)​=elimx→∞​1+2x2​=e0=1