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→0​xex−1​=limx→0​x2ex(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→0​x2ex(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→0​x2ex(x2+x−1)+1​=limx→0​2xex(x2+3x)​=23​