31.写出下列函数在\(x=0\)的带佩亚诺余项的泰勒公式;
(1)\(\cos x^2\);
\(\cos x=\sum\limits_{k=0}^n(-1)^k\frac{x^{2k}}{(2k)!}+o(x^{2n}).\)
\(\cos x^2=\sum\limits_{k=0}^n(-1)^k\frac{x^{4k}}{(2k)!}+o(x^{4n})\).
(2)\(\sin^3x\);
\(\sin x=\sum\limits_{k=0}^n(-1)^k\frac{x^{2k+1}}{(2k+1)!}+o(x^{2n+1})\).
\(\sin^3x=\frac{3\sin x-\sin3x}{4}=\sum\limits_{k=0}^n(-1)^k(\frac{3}{4}-\frac{3^{2k+1}}{4})\frac{x^{2k+1}}{(2k+1)!}+o(x^{2n+1})\).
(3)\(\frac{1}{(1+x)^2}\);
\(f(x)=\frac{1}{(1+x)^2},f^{(n)}(x)=(-1)^n\frac{(n+1)!}{(1+x)^{n+2}}\).
\(\frac{1}{(1+x)^2}=\sum\limits_{k=0}^n(-1)^k(k+1)x^k+o(x^n)\)
(4)\(\frac{x^3+2x+1}{x-1}\);
\(\frac{x^3+2x+1}{x-1}=x^2+x+3+\frac{4}{x-1}=x^2+x+3-4\sum\limits_{k=0}^nx^k+o(x^n)=-3x^2-3x-1-4\sum\limits_{k=3}^nx^k+o(x^n)\).
32.写出下列函数在\(x=0\)的带有佩亚诺余项的泰勒公式,要求至所指定的阶数:
(1)\(\frac{x}{\sin x}\qquad(x^4)\);
\(\frac{x}{\sin x}=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+o(x^4)\).
\((a_0+a_2x^2+a_4x^4+o(x^4))(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^5))=a_0x+(a_2-\frac{x_0}{6})x^3+(a_4-\frac{a_2}{6}+\frac{a_0}{120})x^5+o(x^5)=x\).
\(a_0=1,a_2=\frac{1}{6},a_4=\frac{7}{360}\).
(2)\(\ln(\cos x+\sin x)\qquad(x^4)\);
\(\ln(1+\sin2x)=2x-2x^2+\frac{4x^3}{3}-\frac{4x^4}{3}+o(x^4)\).
\(\ln(\cos x+\sin x)=x-x^2+\frac{2x^3}{3}-\frac{2x^4}{3}+o(x^4)\).
(3)\(\frac{x}{2x^2+x-1}\qquad(x^3)\);
\(\frac{1}{2x^2+x-1}=-1-x-3x+o(x^2)\).
\(\frac{x}{2x^2+x-1}=-x-x^2-3x^2+o(x^3)\).
(4)\(\ln\frac{1+x}{1-2x}\qquad(x^n)\);
\(\ln(1+x)=\sum\limits_{k=1}^n(-1)^{k+1}\frac{x^k}{k}+o(x^n)\).
\(\ln(1-2x)=\sum\limits_{k=1}^n2^k\frac{x^k}{k}+o(x^n)\).
\(\ln\frac{1+x}{1-2x}=\sum\limits_{k=1}^n((-1)^{k+1}+2^k)\frac{x^k}{k}+o(x^n)\).
(5)\(\ln(1+x+x^2+x^3)\qquad(x^6)\);
\(\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\frac{x^6}{6}+o(x^6)\).
\(\ln(1-x^4)=-x^4+o(x^6)\).
\(\ln(1+x+x^2+x^3)=x+\frac{x^2}{2}+\frac{x^3}{3}-\frac{3x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+o(x^6)\).
(6)\(\frac{1+x+x^2}{1-x+x^2}\qquad(x^4)\).
\(\frac{1}{1-x+x^2}=1+x-x^3-x^4+o(x^4)\).
\(\frac{1+x+x^2}{1-x+x^2}=1+2x+2x^2-2x^4+o(x^4)\).
33.确定常数\(a,b\),使当\(x\rightarrow0\)时,
(1)\(f(x)=(a+b\cos x)\sin x-x\)为\(x\)的\(5\)阶无穷小量;
\(f(x)=(a+b-\frac{bx^2}{2}+\frac{bx^4}{24}+o(x^4))(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^5))-x\)
\(=(a+b-1)x-(\frac{a}{6}+\frac{2b}{3})x^3+(\frac{a}{120}-\frac{b}{30})x^5\).
\(a=\frac{4}{3},b=-\frac{1}{3}\).
(2)\(f(x)=\text{e}^x-\frac{1+ax}{1-bx}\)是\(x\)的\(3\)阶无穷小量.
\(f(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)-(1+ax)(1+bx+b^2x^2+b^3x^3+o(x^3))\)
\(=(1-a-b)x+(\frac{1}{2}-ab-b^2)x^2+(\frac{1}{6}-ab^2-b^3)x^3+o(x^3)\).
\(a=\frac{1}{2},b=\frac{1}{2}\).
34.求下列极限
(1)\(\lim\limits_{x\rightarrow0}(\frac{1}{x}-\frac{1}{\sin x})\);
解 \(\lim\limits_{x\rightarrow0}(\frac{1}{x}-\frac{1}{\sin x})=\lim\limits_{x\rightarrow0}\frac{\sin x=x}{x\sin x}=\lim\limits_{x\rightarrow0}\frac{x+o(x^2)-x}{x(x+o(x))}=0\).
(2)\(\lim\limits_{x\rightarrow0}\frac{e^{x^3}-1-x^3}{sin^62x}\);
解 \(\lim\limits_{x\rightarrow0}\frac{1+x^3+\frac{x^6}{2}+o(x^6)-1-x^3}{64x^6+o(x^6)}=\frac{1}{128}\).
(3)\(\lim\limits_{x\rightarrow+\infty}(\sqrt[3]{x^3-3x}-\sqrt{x^2-2x})\);
解 \(\lim\limits_{x\rightarrow0+0}\frac{\sqrt[3]{1-3x^2}-\sqrt{1-2x}}{x}=\lim\limits_{x\rightarrow0+0}\frac{o(x^2)+x+o(x^2)}{x}=1\).
(4)\(\lim\limits_{x\rightarrow\infty}(x+\frac{1}{2})\ln(1+\frac{1}{x})\);
解 \(\lim\limits_{x\rightarrow\infty}(x+\frac{1}{2})\ln(1+\frac{1}{x})=\lim\limits_{x\rightarrow0}\frac{(x+2)\ln(1+x)}{2x}=\lim\limits_{x\rightarrow0}\frac{(x+2)(x+o(x^2))}{2x}=1\).
(5)\(\lim\limits_{x\rightarrow0}(\frac{\tan x}{x})^{\frac{1}{x^2}}\);
解 \(\lim\limits_{x\rightarrow0}(\frac{\tan x}{x})^{\frac{1}{x^2}}=\exp\lim\limits_{x\rightarrow0}(\frac{\ln\frac{\tan x}{x}}{x^2})=\exp(\lim\limits_{x\rightarrow0}\frac{\tan x-x}{x^3})=\text{e}^{\frac{1}{3}}\).
(6)\(\lim\limits_{x\rightarrow\infty}x^2\ln(x\sin\frac{1}{x})\).
解 \(\lim\limits_{x\rightarrow\infty}x^2\ln(x\sin\frac{1}{x})=\lim\limits_{x\rightarrow0}\frac{\ln(\frac{\sin x}{x})}{x^2}=\lim\limits_{x\rightarrow0}\frac{\sin x-x}{x^2}=-\frac{1}{6}\).
