首页工科数学分析函数的导数

# 洛必达法则 1（00\frac 0 0 型）

定理 11：设 f,gf, g 在区间 (x0,x0+δ)(x_0, x_0 + \delta) 有定义，g(x)≠0g(x) \not = 0，满足：

  1. lim⁡x→x0+f(x)=0,lim⁡x→x0+g(x)=0\lim\limits_{x \to x_0^+} f(x) = 0, \lim\limits_{x \to x_0^+} g(x) = 0；

  2. f,gf, g 在区间 (x0,x0+δ)(x_0, x_0 + \delta) 可导，且 g′(x)≠0g'(x) \not = 0；

  3. lim⁡x→x0+f′(x)g′(x)=a\lim\limits_{x \to x_0^+} \dfrac {f'(x)} {g'(x)} = a（aa 为有限或无穷大）。

则有 lim⁡x→x0+f(x)g(x)=lim⁡x→x0+f′(x)g′(x)=a\lim\limits_{x \to x_0^+} \dfrac {f(x)} {g(x)} = \lim\limits_{x \to x_0^+} \dfrac {f'(x)} {g'(x)} = a。

证明：补充定义：

F(x)={f(x),x∈(x0,x0+δ)0,x=x0,G(x)={g(x),x∈(x0,x0+δ)0,x=x0F(x) = \begin {cases} f(x), & x \in (x_0, x_0 + \delta) \\ 0, & x = x_0 \end {cases}, G(x) = \begin {cases} g(x), & x \in (x_0, x_0 + \delta) \\ 0, & x = x_0 \end {cases}

则对 ∀ x∈(x0,x0+δ)\forall \, x \in (x_0, x_0 + \delta)，F,GF, G 在 [x0,x][x_0, x] 连续，在 (x0,x)(x_0, x) 内可导。

由柯西中值定理：∃ ξ∈(x0,x)\exist \, \xi \in (x_0, x)，使得：

f(x)g(x)=F(x)G(x)=F(x)−F(x0)G(x)−G(x0)=F′(ξ)G′(ξ)=f′(ξ)g′(ξ)\frac {f(x)} {g(x)} = \frac {F(x)} {G(x)} = \frac {F(x) - F(x_0)} {G(x) - G(x_0)} = \frac {F'(\xi)} {G'(\xi)} = \frac {f'(\xi)} {g'(\xi)}

而当 x→x0+x \to x_0^+ 时，ξ→x0+\xi \to x_0^+，因此：

lim⁡x→x0+f(x)g(x)=lim⁡ξ→x0+f′(ξ)g′(ξ)=lim⁡x→x0+f′(x)g′(x)=a\lim_{x \to x_0^+} \frac {f(x)} {g(x)} = \lim_{\xi \to x_0^+} \frac {f'(\xi)} {g'(\xi)} = \lim_{x \to x_0^+} \frac {f'(x)} {g'(x)} = a

说明：对 x→x0−,x→x0,x→∞,x→±∞x \to x_0^-, x \to x_0, x \to \infty, x \to \pm \infty 也成立。

定理 22：设 f,gf, g 在 (a,+∞)(a, + \infty) 内有定义，g(x)≠0g(x) \not = 0，且 lim⁡x→+∞f(x)=0,lim⁡x→+∞g(x)=0,g′(x)≠0\lim\limits_{x \to + \infty} f(x) = 0, \lim\limits_{x \to + \infty} g(x) = 0, g'(x) \not = 0，若 lim⁡x→+∞f′(x)g′(x)=A\lim\limits_{x \to + \infty} \dfrac {f'(x)} {g'(x)} = A（AA 可为有限或无穷），则 lim⁡x→+∞f(x)g(x)=A\lim\limits_{x \to + \infty} \dfrac {f(x)} {g(x)} = A。

证明：令 t=1xt = \dfrac 1 x，并利用 定理 11 即可。

lim⁡x→+∞f(x)g(x)=lim⁡t→0+f(1t)g(1t)=lim⁡t→0+f′(1t)(−1t2)g′(1t)(−1t2)=lim⁡t→0+f′(1t)g′(1t)=lim⁡x→+∞f′(x)g′(x)\lim_{x \to + \infty} \frac {f(x)} {g(x)} = \lim_{t \to 0^+} \frac {f(\frac 1 t)} {g(\frac 1 t)} = \lim_{t \to 0^+} \frac {f'(\frac 1 t)(- \frac 1 {t^2})} {g'(\frac 1 t)(- \frac 1 {t^2})} = \lim_{t \to 0^+} \frac {f'(\frac 1 t)} {g'(\frac 1 t)} = \lim_{x \to +\infty} \frac {f'(x)} {g'(x)}

# 洛必达法则 2（∞∞\frac {\infty} {\infty} 型）

定理 33：设 f,gf, g 在内满足：

  1. lim⁡x→x0+g(x)=∞\lim\limits_{x \to x_0^+} g(x) = \infty；

  2. f,gf, g 在区间 (x0,x0+δ)(x_0, x_0 + \delta) 可导，且 g′(x)≠0g'(x) \not = 0；

  3. lim⁡x→x0+f′(x)g′(x)=l\lim\limits_{x \to x_0^+} \dfrac {f'(x)} {g'(x)} = l（ll 为有限或无穷大）。

则有 lim⁡x→x0+f(x)g(x)=lim⁡x→x0+f′(x)g′(x)=l\lim\limits_{x \to x_0^+} \dfrac {f(x)} {g(x)} = \lim\limits_{x \to x_0^+} \dfrac {f'(x)} {g'(x)} = l。

说明：

  1. 并没要求 lim⁡x→x0+f(x)=∞\lim\limits_{x \to x_0^+} f(x) = \infty

  2. 可推广到 x→x0−,x→x0,x→±∞,x→∞x \to x_0^-, x \to x_0, x \to \pm \infty, x \to \infty

注意：

  1. 可多次应用洛必达法则来求极限，每次 应用前都要验证条件；

  2. 与 各种方法综合使用（提出常用因子，等价代换，变量替换）可以简化运算。

  3. 注意条件 lim⁡x→x0+f′(x)g′(x)\lim\limits_{x \to x_0^+} \dfrac {f'(x)} {g'(x)} 要存在或为无穷大！（否则要用其他方法）例如：

    lim⁡x→+∞x+sin⁡xx=lim⁡x→∞1+cos⁡x1\lim\limits_{x \to + \infty} \dfrac {x + \sin x} {x} = \lim\limits_{x \to \infty} \dfrac {1 + \cos x} {1} 不存在，不能洛必达！事实上，lim⁡x→∞x+sin⁡xx=lim⁡x→∞(1+sin⁡xx)=1\lim\limits_{x \to \infty} \dfrac {x + \sin x} {x} = \lim\limits_{x \to \infty} (1 + \dfrac {\sin x} {x}) = 1。

例 1.1. 求下列极限：

lim⁡x→0+x−sin⁡xxx\lim\limits_{x \to 0^+} \dfrac {\sqrt x - \sin \sqrt x} {x \sqrt x}

解：令 y=xy = \sqrt x，则有：

lim⁡x→0+x−sin⁡xxx=lim⁡y→0+y−sin⁡yy3=lim⁡y→0+1−cos⁡y3y2=16\lim\limits_{x \to 0^+} \dfrac {\sqrt x - \sin \sqrt x} {x \sqrt x} = \lim_{y \to 0^+} \frac {y - \sin y} {y^3} = \lim_{y \to 0^+} \frac {1 - \cos y} {3y^2} = \frac 1 6

# 其他不定型

关键：将其他类型未定式化为洛必达法则可解决的类型。

# 0⋅∞0 \cdot \infty 型

步骤：0⋅∞⇒1∞⋅∞0 \cdot \infty \Rightarrow \dfrac 1 \infty \cdot \infty，或 0⋅∞⇒0⋅100 \cdot \infty \Rightarrow 0 \cdot \dfrac 1 0

# ∞−∞\infty - \infty 型

步骤：∞−∞⇒10−10⇒0−00⋅0\infty - \infty \Rightarrow \dfrac 1 0 - \dfrac 1 0 \Rightarrow \dfrac {0 - 0} {0 \cdot 0}。

# 00,1∞,∞00^0, 1^\infty, \infty^0 型

步骤：

001∞∞0}→取对数{0⋅ln⁡0∞⋅ln⁡10⋅ln⁡∞⇒0⋅∞\begin {rcases} 0^0 \\ 1^\infty \\ \infty^0 \end {rcases} \xrightarrow {取对数} \begin {cases} 0 \cdot \ln 0 \\ \infty \cdot \ln 1 \\ 0 \cdot \ln \infty \end {cases} \Rightarrow 0 \cdot \infty

# 习题

  1. 计算下列极限：

    (1) lim⁡x→0x−sin⁡xx3\lim\limits_{x \to 0} \dfrac {x - \sin x} {x^3}

    (2) lim⁡x→0x−x22−ln⁡(1+x)x3\lim\limits_{x \to 0} \dfrac {x - \dfrac {x^2} 2 - \ln (1 + x)} {x^3}

    (3) lim⁡x→1ln⁡[cos⁡(x−1)]1−sin⁡πx2\lim\limits_{x \to 1} \dfrac {\ln [\cos (x - 1)]} {1 - \sin \dfrac {\pi x} 2}

    (4) lim⁡x→0ex−1+2xln⁡(1+x2)\lim\limits_{x \to 0} \dfrac {e^x - \sqrt {1 + 2x}} {\ln (1 + x^2)}

    (5) lim⁡x→0x2e1x2\lim\limits_{x \to 0} x^2 e^{\frac 1 {x^2}}

    (6) lim⁡x→0(1sin⁡x−1x)\lim\limits_{x \to 0} \left( \dfrac 1 {\sin x} - \dfrac 1 x \right)

    (7) lim⁡x→0(1x−1ex−1)\lim\limits_{x \to 0} \left( \dfrac 1 x - \dfrac 1 {e^x - 1} \right)

    (8) lim⁡x→0(tan⁡xx)1x2\lim\limits_{x \to 0} \left( \dfrac {\tan x} x \right)^{\frac 1 {x^2}}

    (9) lim⁡x→0(ln⁡1x)x\lim\limits_{x \to 0} \left( \ln \dfrac 1 x \right)^x

    (10) lim⁡x→0(tan⁡x)sin⁡x\lim\limits_{x \to 0} (\tan x)^{\sin x}

    (11) lim⁡x→∞(tan⁡πx2x+1)1x\lim\limits_{x \to \infty} \left( \tan \dfrac {\pi x} {2x + 1} \right)^{\frac 1 x}

    (1) 解：

    lim⁡x→0x−sin⁡xx3=lim⁡x→01−cos⁡x3x2=lim⁡x→012x23x2=16 \lim\limits_{x \to 0} \dfrac {x - \sin x} {x^3} = \lim_{x \to 0} \frac {1 - \cos x} {3x^2} = \lim_{x \to 0} \frac {\frac 1 2 x^2} {3 x^2} = \frac 1 6

    (2) 解：

    lim⁡x→0x−x22−ln⁡(1+x)x3=lim⁡x→01−x−11+x3x2=lim⁡x→0−1+1(1+x)26x=−lim⁡x→0x+26(1+x)2=−13 \lim\limits_{x \to 0} \dfrac {x - \dfrac {x^2} 2 - \ln (1 + x)} {x^3} = \lim_{x \to 0} \dfrac {1 - x - \dfrac 1 {1 + x}} {3x^2} = \lim_{x \to 0} \frac {-1 + \dfrac 1 {(1 + x)^2}} {6x} = - \lim_{x \to 0} \frac {x + 2} {6(1 + x)^2} = - \frac 1 3

    (3) 解：

    lim⁡x→1ln⁡[cos⁡(x−1)]1−sin⁡πx2=lim⁡x→1cos⁡(x−1)−11−sin⁡πx2=lim⁡x→1−sin⁡(x−1)−π2cos⁡πx2=lim⁡x→1x−1π2cos⁡πx2=lim⁡x→11−π24sin⁡πx2=−4π2 \lim\limits_{x \to 1} \dfrac {\ln [\cos (x - 1)]} {1 - \sin \dfrac {\pi x} 2} = \lim_{x \to 1} \frac {\cos (x - 1) - 1} {1 - \sin \dfrac {\pi x} 2} = \lim_{x \to 1} \frac {- \sin (x - 1)} {- \dfrac \pi 2 \cos \dfrac {\pi x} 2} = \lim_{x \to 1} \frac {x - 1} {\dfrac \pi 2 \cos \dfrac {\pi x} 2} = \lim_{x \to 1} \frac {1} {- \dfrac {\pi^2} 4 \sin \dfrac {\pi x} 2} = - \frac 4 {\pi^2}

    (4) 解：

    lim⁡x→0ex−1+2xln⁡(1+x2)=lim⁡x→0ex−1+2xx2 \lim_{x \to 0} \frac {e^x - \sqrt {1 + 2x}} {\ln (1 + x^2)} = \lim_{x \to 0} \frac {e^x - \sqrt {1 + 2x}} {x^2}

    令 t=1+2x→1t = \sqrt {1 + 2x} \to 1，则 x=t2−12x = \dfrac {t^2 - 1} 2，代入得：

    lim⁡x→0ex−1+2xx2=lim⁡t→1et2−12−t(t2−1)24=lim⁡t→1tet2−12−1(t2−1)t=lim⁡t→1(1+t2)et2−123t2−1=1 \lim_{x \to 0} \frac {e^x - \sqrt {1 + 2x}} {x^2} = \lim_{t \to 1} \frac {e^{\frac {t^2 - 1} 2} - t} {\frac {(t^2 - 1)^2} 4} = \lim_{t \to 1} \frac {t e^{\frac {t^2 - 1} 2} - 1} {(t^2 - 1)t} = \lim_{t \to 1} \frac {(1 + t^2) e^{\frac {t^2 - 1} 2}} {3t^2 - 1} = 1

    (5) 解：

    lim⁡x→0x2e1x2=lim⁡x→0e1x21x2=lim⁡x→0e1x2(−2x3)−2x3=lim⁡x→0e1x2=+∞ \lim_{x \to 0} x^2 e^{\frac 1 {x^2}} = \lim_{x \to 0} \frac {e^{\frac 1 {x^2}}} {\frac 1 {x^2}} = \lim_{x \to 0} \frac {e^{\frac 1 {x^2}} (- \frac 2 {x^3})} {- \frac 2 {x^3}} = \lim_{x \to 0} e^{\frac 1 {x^2}} = + \infty

    (6) 解：

    lim⁡x→0(1sin⁡x−1x)=lim⁡x→0x−sin⁡xx2=lim⁡x→01−cos⁡x2x=lim⁡x→012x22x=lim⁡x→0x4=0 \lim_{x \to 0} \left( \frac 1 {\sin x} - \frac 1 x \right) = \lim_{x \to 0} \frac {x - \sin x} {x^2} = \lim_{x \to 0} \frac {1 - \cos x} {2x} = \lim_{x \to 0} \frac {\frac 1 2 x^2} {2x} = \lim_{x \to 0} \frac x 4 = 0

    (7) 解：

    lim⁡x→0(1x−1ex−1)=lim⁡x→0ex−1−xx(ex−1)=lim⁡x→0ex−1−xx2=lim⁡x→0ex−12x=lim⁡x→0x2x=12 \lim_{x \to 0} \left( \frac 1 x - \frac 1 {e^x - 1} \right) = \lim_{x \to 0} \frac {e^x - 1 - x} {x (e^x - 1)} = \lim_{x \to 0} \frac {e^x - 1 - x} {x^2} = \lim_{x \to 0} \frac {e^x - 1} {2x} = \lim_{x \to 0} \frac {x} {2x} = \frac 1 2

    (8) 解：

    lim⁡x→0tan⁡xx−1x2=lim⁡x→0tan⁡x−xx3=lim⁡x→0sec⁡2x−13x2=lim⁡x→0tan⁡2x3x2=13lim⁡x→0(tan⁡xx)1x2=lim⁡x→0etan⁡xx−1x2=e13 \lim_{x \to 0} \frac {\dfrac {\tan x} x - 1} {x^2} = \lim_{x \to 0} \frac {\tan x - x} {x^3} = \lim_{x \to 0} \frac {\sec^2 x - 1} {3x^2} = \lim_{x \to 0} \frac {\tan^2 x} {3x^2} = \frac 1 3 \\ \lim_{x \to 0} \left( \frac {\tan x} x \right)^{\frac 1 {x^2}} = \lim_{x \to 0} e^{\frac {\frac {\tan x} x - 1} {x^2}} = e^{\frac 1 3}

    (9) 解：

    lim⁡x→0xln⁡(ln⁡x)=lim⁡x→0ln⁡(ln⁡1x)1x=lim⁡x→01ln⁡1xx(−1x2)(−1x2)=lim⁡x→0xln⁡1x=0lim⁡x→0(ln⁡1x)x=lim⁡x→0exln⁡(ln⁡x)=1 \lim_{x \to 0} x \ln (\ln x) = \lim_{x \to 0} \frac {\ln \left( \ln \dfrac 1 x \right)} {\dfrac 1 x} = \lim_{x \to 0} \frac {\dfrac 1 {\ln \frac 1 x} x \left( - \dfrac 1 {x^2} \right)} {\left( - \dfrac 1 {x^2} \right)} = \lim_{x \to 0} \frac x {\ln \dfrac 1 x} = 0 \\ \lim_{x \to 0} \left( \ln \frac 1 x \right)^x = \lim_{x \to 0} e^{x \ln (\ln x)} = 1

    (10) 解：

    lim⁡x→0sin⁡xln⁡(tan⁡x)=lim⁡x→0ln⁡(tan⁡x)1sin⁡x=lim⁡x→0sec⁡2xtan⁡x−csc⁡xcot⁡x=lim⁡x→0−sec⁡2xcsc⁡x=−lim⁡x→0sin⁡xcos⁡2x=0lim⁡x→0(tan⁡x)sin⁡x=lim⁡x→0esin⁡xln⁡(tan⁡x)=1 \lim_{x \to 0} \sin x \ln (\tan x) = \lim_{x \to 0} \frac {\ln (\tan x)} {\dfrac 1 {\sin x}} = \lim_{x \to 0} \frac {\dfrac {\sec^2 x} {\tan x}} {- \csc x \cot x} = \lim_{x \to 0} - \frac {\sec^2 x} {\csc x} = - \lim_{x \to 0} \frac {\sin x} {\cos^2 x} = 0 \\ \lim_{x \to 0} (\tan x)^{\sin x} = \lim_{x \to 0} e^{\sin x \ln (\tan x)} = 1

    (11) 解：令 t=πx2x+1→π2t = \dfrac {\pi x} {2x + 1} \to \dfrac \pi 2，则：

    lim⁡x→∞ln⁡tan⁡πx2x+1x=lim⁡t→π2ln⁡tan⁡ttπ−2t=lim⁡t→π2sec⁡2ttan⁡tπ(π−2t)2=lim⁡t→π2(π−2t)2πsin⁡tcos⁡t=lim⁡t→π2(π−2t)2πcos⁡t=lim⁡t→pi24(π−2t)πsin⁡t=0lim⁡x→∞(tan⁡πx2x+1)1x=lim⁡x→∞eln⁡tan⁡πx2x+1x=1 \lim_{x \to \infty} \frac {\ln \tan \dfrac {\pi x} {2x + 1}} {x} = \lim_{t \to \frac \pi 2} \frac {\ln \tan t} {\dfrac t {\pi - 2t}} = \lim_{t \to \frac \pi 2} \frac {\dfrac {\sec^2 t} {\tan t}} {\dfrac \pi {(\pi - 2t)^2}} = \lim_{t \to \frac \pi 2} \frac {(\pi - 2t)^2} {\pi \sin t \cos t} = \lim_{t \to \frac \pi 2} \frac {(\pi - 2t)^2} {\pi \cos t} = \lim_{t \to \frac pi 2} \frac {4(\pi - 2t)} {\pi \sin t} = 0 \\ \lim_{x \to \infty} \left( \tan \frac {\pi x} {2x + 1} \right)^{\frac 1 x} = \lim_{x \to \infty} e^{\frac {\ln \tan \frac {\pi x} {2x + 1}} {x}} = 1

  2. 已知 f(x)={g(x)x,x≠00,x=0f(x) = \begin {cases} \dfrac {g(x)} x, & x \not = 0 \\ 0, & x = 0 \end {cases}，其中 g(0)=0,g′(0)=0,g′′(0)=Ag(0) = 0, g'(0) = 0, g''(0) = A，求 f′(0)f'(0)。

    解：

    f′(0)=lim⁡x→0g(x)xx=lim⁡x→0g(x)x2=lim⁡x→0g′(x)2x=lim⁡x→0g′′(x)2=A2 \begin {aligned} f'(0) &= \lim_{x \to 0} \frac {\dfrac {g(x)} x} {x} = \lim_{x \to 0} \frac {g(x)} {x^2} = \lim_{x \to 0} \frac {g'(x)} {2x} = \lim_{x \to 0} \frac {g''(x)} 2 = \frac A 2 \end {aligned}

  3. 设 f(x)f(x) 在点 xx 处有二阶导数，求证：f′′(x)=lim⁡h→0f(x+h)+f(x−h)−2f(x)h2f''(x) = \lim\limits_{h \to 0} \dfrac {f(x + h) + f(x - h) - 2 f(x)} {h^2}。由此推出结论：若 f(x)f(x) 是二阶可导的凸函数，必有 f′′(x)≥0f''(x) \ge 0。

    证明：

    lim⁡h→0f(x+h)+f(x−h)−2f(x)h2=lim⁡h→0f(x+h)−f(x)h−f(x)−f(x−h)hh=lim⁡h→0f′(x)−f′(x−h)h=f′′(x−h)=f′′(x) \lim_{h \to 0} \frac {f(x + h) + f(x - h) - 2 f(x)} {h^2} = \lim_{h \to 0} \frac {\dfrac {f(x + h) - f(x)} {h} - \dfrac {f(x) - f(x - h)} {h}} {h} = \lim_{h \to 0} \frac {f'(x) - f'(x - h)} {h} = f''(x - h) = f''(x)

    由于 f(x)f(x) 是二阶可导的凸函数，由凸函数性质可知，f(x+h)+f(x−h)≥2f(x)f(x + h) + f(x - h) \ge 2 f(x)，从而证得 f′′(x)≥0f''(x) \ge 0。

  4. 设 x1=1,xn+1=ln⁡(1+xn),n=1,2,⋯x_1 = 1, x_{n + 1} = \ln (1 + x_n), n = 1, 2, \cdots，求极限 lim⁡n→∞nxn\lim\limits_{n \to \infty} n x_n。

    解：已知 1≥x1>01 \ge x_1 > 0，假设 1≥xn>01 \ge x_n > 0，则：

    xn+1−xn=ln⁡(1+xn)−xn<0 x_{n + 1} - x_n = \ln (1 + x_n) - x_n < 0

    即数列 {xn}\{ x_n \} 单调递减。又由于当 xn>0x_n > 0 时，xn+1=ln⁡(1+xn)>0x_{n + 1} = \ln (1 + x_n) > 0，因此 00 是数列 {xn}\{ x_n \} 的一个下界，因此数列 {xn}\{ x_n \} 收敛，设 lim⁡n→∞xn=A\lim\limits_{n \to \infty} x_n = A，对递推式两边取极限：

    A=ln⁡(1+A) A = \ln (1 + A)

    解得 A=0A = 0。由斯图尔兹定理：

    lim⁡n→∞nan=lim⁡n→∞n1an=lim⁡n→∞11an+1−1an=lim⁡n→∞anan+1an−an+1=lim⁡n→∞anln⁡(1+an)an−ln⁡(1+an) \lim_{n \to \infty} n a_n = \lim_{n \to \infty} \frac n {\frac 1 {a_n}} = \lim_{n \to \infty} \frac 1 {\frac 1 {a_{n + 1}} - \frac 1 {a_n}} = \lim_{n \to \infty} \frac {a_n a_{n + 1}} {a_n - a_{n + 1}} = \lim_{n \to \infty} \frac {a_n \ln (1 + a_n)} {a_n - \ln (1 + a_n)}

    设函数 f(x)=x−ln⁡(1+x)xln⁡(1+x)f(x) = \dfrac {x - \ln (1 + x)} {x \ln (1 + x)}，由洛必达法则：

    lim⁡x→0x−ln⁡(1+x)xln⁡(1+x)=lim⁡x→01−11+xln⁡(1+x)+x1+x=lim⁡x→0x(1+x)ln⁡(1+x)+x=lim⁡x→012+ln⁡(1+x)=12 \lim_{x \to 0} \frac {x - \ln (1 + x)} {x \ln (1 + x)} = \lim_{x \to 0} \frac {1 - \dfrac 1 {1 + x}} {\ln (1 + x) + \dfrac x {1 + x}} = \lim_{x \to 0} \frac x {(1 + x) \ln (1 + x) + x} = \lim_{x \to 0} \frac {1} {2 + \ln (1 + x)} = \frac 1 2

    由海涅定理可知，lim⁡n→∞anln⁡(1+an)an−ln⁡(1+an)=2\lim\limits_{n \to \infty} \dfrac {a_n \ln (1 + a_n)} {a_n - \ln (1 + a_n)} = 2，即 lim⁡n→∞nxn=2\lim\limits_{n \to \infty} n x_n = 2
    没想到利用海涅定理将数列极限转化为函数极限再用洛必达求出函数极限