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兩種重要極限混合的極限題

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自然指數的定義

e:= lim⁡n→∞(1+1n)n= lim⁡x→0(1+x)1x\begin{align} e:=&\,\lim_{n\to\infty}\big(1+\frac{1}{n}\big)^{n}\\[4mm] =&\,\lim_{x\to0}\big(1+x\big)^{\frac{1}{x}} \end{align}

複習與回顧

續上一篇討論重要極限 lim⁡x→0sin⁡xx\displaystyle\lim_{x\to0}\frac{\sin x}{x}的幾個延伸題，
這回繼續討論由自然指數定義衍生出來的重要極限，及其相關的極限題。

回顧書中出現的極限

在《白話微積分》中，列出了幾個與 ee 相關的重要極限：

  1. lim⁡n→∞(1+1n)n=e\displaystyle\lim_{n\to\infty}\big(1+\frac{1}{n}\big)^n=e

  2. lim⁡n→∞(1+an)n=ea\displaystyle\lim_{n\to\infty}\big(1+\frac{a}{n}\big)^n=e^a

  3. lim⁡x→0(1+x)1x=e\displaystyle\lim_{x\to0}\big(1+x\big)^{\frac{1}{x}}=e

  4. lim⁡x→0(1+ax)1x=ea\displaystyle\lim_{x\to0}\big(1+ax\big)^{\frac{1}{x}}=e^a

  5. lim⁡x→0ln⁡(1+x)x=1\displaystyle\lim_{x\to0}\frac{\ln\big(1+x\big)}{x}=1

  6. lim⁡x→0ex−1x=1\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1

  7. lim⁡x→0ax−1x=ln⁡a\displaystyle\lim_{x\to0}\frac{a^x-1}{x}=\ln a

回顧前文

也别忘了上篇文章所見過的幾個常用已知極限

  1. lim⁡x→0sin⁡axax=1\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=1

  2. lim⁡x→0tan⁡axax=1\displaystyle\lim_{x\to0}\frac{\tan ax}{ax}=1

  3. lim⁡x→01−cos⁡xx2=12\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}

回顧書中例題

\begin{align*} (1) \quad&\; \lim_{x\to0}\frac{5^x-3^x}{x}\\[4mm] =&\,\lim_{x\to0}\frac{\big(5^x-1\big)-\big(3^x-1\big)}{x}\\[4mm] =&\,\lim_{x\to0} \frac{5^x-1}{x}-\lim_{x\to0}\frac{3^x-1}{x}\\[4mm] =&\,\ln5-\ln3=\ln\frac{5}{3}\\[4mm] (2) \quad&\; \lim_{x\to0}\big(1+2x\big)^{\frac{1}{\sin(x)}}\\[4mm] =&\,\lim_{x\to0}\bigg\[ \big(1+2x\big)^{\frac{1}{2x}} \bigg\]^{\frac{x}{\sin(x)}\cdot2} =e^2\\[4mm] (3) \quad&\; \lim_{x\to0} \big(1+x^2\big)^{\cot^2(x)}\\[4mm] =&\,\lim_{x\to0}\bigg\[ \big(1+x^2\big)^{\frac{1}{x^2}} \bigg\]^{x^2\cdot\cot^2(x)}\\[4mm] =&\,\lim_{x\to0}\bigg\[ \big(1+x^2\big)^{\frac{1}{x^2}} \bigg\]^{\frac{x^2}{\sin^2(x)}\cdot\cos^2(x)}\\[4mm] =&\,e^{(1\cdot1)}=e\\[4mm] (4) \quad&\; \lim_{x\to0} \big(e^x+\sin(x)\big)^{\frac{2}{x}}\\[4mm] =&\,\lim_{x\to0}e^2\cdot \Big(1+\frac{\sin(x)}{e^x} \Big)^{\frac{2}{x}}\\[4mm] =&\,e^2\cdot\lim_{x\to0}\bigg\[ \big(1+\frac{\sin(x)}{e^x}\big)^{\frac{e^x}{\sin(x)}} \bigg\]^{\frac{\sin(x)}{x} \cdot2\cdot\frac{1}{e^x}}\\[4mm] =&\,e^2\cdot e^2=e^4\\[4mm] (5) \quad&\; \lim_{x\to\infty} \Big(\frac{x+a}{x-a}\Big)^{x}\\[4mm] =&\,\lim_{x\to\infty} \Big(\frac{x-a+2a}{x-a}\Big)^{x}\\[4mm] =&\,\lim_{x\to\infty} \Big(1+\frac{2a}{x-a}\Big)^{x-a} \cdot\Big(1+\frac{2a}{x-a}\Big)^{7a}\\[4mm] =&\,e^{2a}\cdot1=e^3\;\; \Rightarrow a=\frac{3}{2} \end{align*}

延伸題

延伸題 1

lim⁡x→0(1+3x)2sin⁡x\begin{align*} \lim_{x\to0} \big(1+3x\big)^{\frac{2}{\sin x}} \end{align*}

解

\begin{align} &\,\lim_{x\to0} \big(1+3x\big)^{\frac{2}{\sin x}}\\[4mm] =&\,\lim_{x\to0} \bigg\[\big(1+3x\big)^{\frac{1}{3x}} \bigg\]^{\frac{2}{\sin x}\times3x} &&\colorbox{Lavender}{先對齊$3x$}\\[4mm] =&\,\lim_{x\to0} \bigg\[\big(1+3x\big)^{\frac{1}{3x}} \bigg\]^{\frac{x}{\sin x}\times2\times3} &&\colorbox{Lavender}{再湊$\frac{\sin x}{x}$}\\[4mm] =&\,e^{1\cdot2\cdot3}=e^6 \end{align}

延伸題 2

lim⁡x→0(1−sin⁡x)xln⁡(1+3x2)\begin{align*} \lim_{x\to0} \big(1-\sin x\big)^{\frac{x}{\ln(1+3x^2)}} \end{align*}

解

看見底是 (1+無窮小量)(1+\colorbox{SkyBlue}{\footnotesize 無窮小量})這種形式，
就想到往 lim⁡x→0(1+x)1x\displaystyle\lim_{x\to0}(1+x)^{\frac{1}{x}}去湊

 lim⁡x→0(1−sin⁡x)xln⁡(1+3x2)= lim⁡x→0((1−sin⁡x)1sin⁡x)sin⁡x ⋅ xln⁡(1+3x2)= lim⁡x→0((1−sin⁡x)1sin⁡x)sin⁡xx⋅3x2ln⁡(1+3x3)⋅13= (e−1)1⋅1⋅13=e−13\begin{align} &\,\lim_{x\to0} \big(1-\sin x\big)^{\frac{x}{\ln(1+3x^2)}}\\[4mm] =&\,\lim_{x\to0} \bigg(\big(1-\sin x\big)^{\frac{1}{\sin x}} \bigg)^{\sin x\,\cdot\,\frac{x}{\ln(1+3x^2)}} \\[4mm] =&\,\lim_{x\to0} \bigg(\big(1-\sin x\big)^{\frac{1}{\sin x} }\bigg)^{\frac{\sin x}{x}\cdot\frac{3x^2}{\ln(1+3x^3)} \cdot\frac{1}{3}}\\[4mm] =&\,\big(e^{-1}\big)^{1\cdot1\cdot\frac{1}{3}} =e^{-\frac{1}{3}} \end{align}

延伸題 3

解

lim⁡x→0xln⁡(1+x)1−cos⁡x\begin{align*} \lim_{x\to0} \frac{x\ln(1+x)}{1-\cos x} \end{align*}

透過觀察題目，不難聯想到這兩已知極限：

 lim⁡x→01−cos⁡xx2=12= lim⁡x→0ln⁡(1+x)x=1\begin{align} &\,\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2} \\[4mm] =&\,\lim_{x\to0}\frac{\ln(1+x)}{x}=1 \end{align}

所以

 lim⁡x→0xln⁡(1+x)1−cos⁡x= lim⁡x→0x1−cos⁡x⋅ln⁡(1+x)先拉開= lim⁡x→0x21−cos⁡x⋅ln⁡(1+x)x再湊= 2⋅1=2\begin{align} &\,\lim_{x\to0} \frac{x\ln(1+x)}{1-\cos x}\\[4mm] =&\,\lim_{x\to0} \frac{x}{1-\cos x} \cdot\ln(1+x) &&\colorbox{Lavender}{先拉開}\\[4mm] =&\,\lim_{x\to0} \frac{x^2}{1-\cos x} \cdot\frac{\ln(1+x)}{x} &&\colorbox{Lavender}{再湊}\\[4mm] =&\,2\cdot1=2 \end{align}

延伸題 4

lim⁡x→0ln⁡(cos⁡x)x2\begin{align*} \lim_{x\to0} \frac{\ln(\cos x)}{x^2} \end{align*}

解

乍看分子的 ln⁡(cos⁡x)\ln(\cos x) 並不滿足我們比較熟悉的
ln⁡(1+無窮小量)\ln(1+\colorbox{SkyBlue}{\footnotesize 無窮小量}) 這種形式，
但我們可以自己湊：

ln⁡(cos⁡x)=ln⁡(1+(cos⁡x−1))\begin{align} \ln(\cos x) =\ln\big(1+(\cos x-1)\big) \end{align}

這樣就有方向了，於是

 lim⁡x→0ln⁡(cos⁡x)x2= lim⁡x→0ln⁡(1+(cos⁡x−1))cos⁡x−1⋅cos⁡x−1x2= 1⋅(−12)=−12\begin{align} &\,\lim_{x\to0} \frac{\ln(\cos x)}{x^2}\\[4mm] =&\,\lim_{x\to0} \frac{\ln\big(1+(\cos x-1)\big)}{\cos x-1} \cdot\frac{\cos x-1}{x^2}\\[4mm] =&\,1\cdot(-\frac{1}{2}) =-\frac{1}{2} \end{align}

延伸題 5

lim⁡x→0(cos⁡x)1ln⁡(1+x2)\begin{align*} \lim_{x\to0} \big(\cos x\big)^{\frac{1}{\ln(1+x^2)}} \end{align*}

解

\begin{align} &\,\lim_{x\to0} \big(\cos x\big)^{\frac{1}{\ln(1+x^2)}}\\[4mm] =&\,\lim_{x\to0}\big(1+ (\cos x-1)\big)^{\frac{1}{\ln(1+x^2)}}\\[4mm] =&\,\lim_{x\to0} \bigg\[\big(1+ (\cos x-1)\big)^{\frac{1}{\cos x-1} }\bigg\]^{\frac{\cos x-1}{\ln(1+x^2)}}\\[4mm] =&\,\lim_{x\to0} \bigg\[\big(1+ (\cos x-1)\big)^{\frac{1}{\cos x-1} }\bigg\]^{\frac{\cos x-1}{x^2} \cdot\frac{x^2}{\ln(1+x^2)}}\\[4mm] =&\,e^{-\frac{1}{2}\cdot1}=e^{-\frac{1}{2}} \end{align}
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