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更新第二三章习题部分

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Didnelpsun
2021-05-31 23:38:13 +08:00
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advanced-math/knowledge/1-function-and-limit/function-and-limit.tex
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@@ -442,15 +442,7 @@ $\therefore \ln\left(1+\dfrac{\tan x}{x}-1\right)\sim\dfrac{\tan x}{x}-1$，$\df
又根据泰勒展开$\tan x-x=x+\dfrac{x^3}{3}+o(x^3)-x-0\cdot x^3=\dfrac{x^3}{3}$。
$\therefore$ \medskip
$e^{\lim\limits_{x\to 0}\frac{1}{x^2}\ln\frac{\tan x}{x}}$
$=e^{\lim\limits_{x\to 0}\frac{1}{x^2}\frac{\tan x-x}{x}}$
$=e^{\lim\limits_{x\to 0}\frac{1}{x^2}\cdot\frac{x^2}{3}}$
$= e^{\frac{1}{3}}$
$\therefore e^{\lim\limits_{x\to 0}\frac{1}{x^2}\ln\frac{\tan x}{x}}=e^{\lim\limits_{x\to 0}\frac{1}{x^2}\frac{\tan x-x}{x}}=e^{\lim\limits_{x\to 0}\frac{1}{x^2}\cdot\frac{x^2}{3}}=e^{\frac{1}{3}}$
根据海涅定理：取$x=\dfrac{1}{n},n\to\infty$，$\lim\limits_{n\to\infty}\left(n\tan\dfrac{1}{n}\right)^{n^2}=e^{\frac{1}{3}}$。
@@ -887,21 +879,9 @@ $\therefore\lim\limits_{x\to 0}\dfrac{\sin x}{x}=1$。
证明：
$\lim\limits_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x$\medskip
$\lim\limits_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=\lim\limits_{x\to\infty}e^{\ln(1+\frac{1}{x})^x}=\lim\limits_{x\to\infty}e^{x\ln(1+\frac{1}{x})}=e^{\lim\limits_{x\to\infty}x\ln(1+\frac{1}{x})}=e^{\lim\limits_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}}$
$=\lim\limits_{x\to\infty}e^{\ln(1+\frac{1}{x})^x}$
$=\lim\limits_{x\to\infty}e^{x\ln(1+\frac{1}{x})}$
$=e^{\lim\limits_{x\to\infty}x\ln(1+\frac{1}{x})}$
$=e^{\lim\limits_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}}$
$=e^{\lim\limits_{x\to\infty}\frac{\left(\frac{1}{1+\frac{1}{x}}\right)\cdot\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}}$
$=e^{\lim\limits_{x\to\infty}\frac{1}{1+x}}$
$=e$
$=e^{\lim\limits_{x\to\infty}\frac{\left(\frac{1}{1+\frac{1}{x}}\right)\cdot\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}}=e^{\lim\limits_{x\to\infty}\frac{1}{1+x}}=e$\medskip
从而$\lim\limits_{\Delta\to\infty}\left(1+\dfrac{1}{\Delta}\right)^\Delta=e$与$\lim\limits_{\Delta\to 0}\left(1+\Delta\right)^{\frac{1}{\Delta}}=e(\Delta\neq 0)$。