错题整理

y=f(x)=ln⁡(x+x2+1)y=f(x)=\ln(x+\sqrt{x^2+1})

求f−1(x)f^{-1}(x)

答

−y=−ln⁡(x+x2+1);=ln⁡1x+x2+1;=ln⁡x−x2+1(x+x2+1)(x−x2+1);=ln⁡(x2+1−x);ey=x2+1+x;e−y=x2+1−x;ey−e−y=2x;y=f−1(x)=ex−e−x2-y = -\ln(x+\sqrt{x^2+1}) ;\\ = \ln{\frac{1}{x+\sqrt{x^2+1}}} ;\\ = \ln{\frac{x-\sqrt{x^2+1}}{(x+\sqrt{x^2+1})(x-\sqrt{x^2+1})}} ;\\ = \ln{(\sqrt{x^2+1}-x)} ;\\ e^y = \sqrt{x^2+1} + x ;\\ e^{-y} = \sqrt{x^2+1} - x ;\\ e^y-e^{-y} = 2x ;\\ y = f^{-1}(x) = \frac{e^x-e^{-x}}{2}

证明f(x)=x1+x2f(x)=\frac{x}{1+x^2}在无穷域下有界

x=0,f(0)=0;x≠0,∣f(x)∣=11∣x∣+∣x∣;1∣x∣+∣x∣≤21∣x∣⋅∣x∣=2;∣f(x)∣≤12x=0, f(0) = 0 ;\\ x \ne 0, |f(x)|=\frac{1}{\frac{1}{|x|} + |x|} ;\\ \frac{1}{|x|} + |x| \le 2\sqrt{\frac{1}{|x|}\cdot|x|} = 2 ;\\ |f(x)| \le \frac{1}{2}

ln⁡x=1,x=e2\ln\sqrt{x}=1, x=e^2

ln⁡x=0,x=1\ln\sqrt{x}=0, x=1

证明 lim⁡n→∞[1+(−1)nn]=1\lim_{n \to \infty}[1+\frac{(-1)^n}{n}]=1

  1. 令N=1ϵ+1N = \frac{1}{\epsilon} + 1
  2. n>N→n>1ϵn>N \to n>\frac{1}{\epsilon}
  3. 即∣1+(−1)nn−1∣<ϵ|1+\frac{(-1)^n}{n}-1| \lt \epsilon
  4. lim⁡n→∞[1+(−1)nn]=1\lim_{n \to \infty}[1+\frac{(-1)^n}{n}]=1

求

a1=a,an+1=12(an+2an);lim⁡n→∞ana_1=a, a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n});\\ \lim_{n \to \infty} a_n

证明

an+1=12(an+2an)≥an⋅2an=2;an+1−an=2−an2an≤0→{an}↓;lim⁡n→∞an=A;lim⁡n→∞an+1=lim⁡n→∞12(an+2an)→A=12(A+2A)→A=2a_{n+1} = \frac{1}{2}(a_n+\frac{2}{a_n}) \ge \sqrt{a_n\cdot\frac{2}{a_n}} = \sqrt{2} ;\\ a_{n+1}-a_n = \frac{2-a_n^2}{a_n}\le 0 \to \{a_n\} \downarrow ;\\ \lim_{n \to \infty} a_n = A ;\\ \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty}\frac{1}{2}(a_n+\frac{2}{a_n}) \to A = \frac{1}{2}(A + \frac{2}{A}) \to A = \sqrt{2}

求

A=lim⁡n→∞∑i=1n1n2+iA = \lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{n^2 + i}}

解

1=lim⁡n→∞nn2+n≤A≤lim⁡n→∞nn2+1=11 = \lim_{n \to \infty} \frac{n}{\sqrt{n^2 + n}} \le A \le \lim_{n \to \infty} \frac{n}{\sqrt{n^2 + 1}} = 1
A=lim⁡n→∞∑i=1nin2+n+iA = \lim_{n\to\infty} \sum_{i=1}^n \frac{i}{n^2+n+i} 12=lim⁡n→∞n(n+1)2(n2+n+n)<A<lim⁡n→∞n(n+1)2(n2+n+1)=12\frac{1}{2} = \lim_{n\to\infty} \frac{n(n+1)}{2(n^2+n+n)} \lt A \lt \lim_{n\to\infty} \frac{n(n+1)}{2(n^2+n+1)} = \frac{1}{2}
an=∑i→nn1n2a_n = \sum_{i \to n}^n \frac{1}{n^2}

证明{an}收敛

an+1−an=1(n+1)2>0→{an}↑;an<1+11⋅2+12⋅3+⋯+1(n−1)⋅n;=1+1−12+12−13+⋯−1n;=2−1n<2a_{n+1} - a_n = \frac{1}{(n+1)^2} > 0 \to \{a_n\} \uparrow ;\\ a_n \lt 1 + \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \cdots + \frac{1}{(n-1)\cdot n} ;\\ = 1 + 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots - \frac{1}{n} ;\\ = 2 - \frac{1}{n} \lt 2

求

a0=0,a1=1,2an+1=an+an−1,;limn→∞ana_0=0,a_1=1, 2a_{n+1}=a_n+a_{n-1}, ;\\ lim_{n \to \infty}a_n an+1−an=(−12)(an−an−1);=(−12)n;an=an−an−1+an−1−⋯+a1−a0+a0;=(−12)n−1+(−12)n−2+⋯(−12)0;=1−(−12)n1−(−12);=A;limn→∞A=23a_{n+1} - a_n = (-\frac{1}{2})(a_n - a_{n-1}) ;\\ = (-\frac{1}{2})^n ;\\ a_n = a_n -a_{n-1} + a_{n-1} - \cdots + a_1 - a_0 + a_0;\\ = (-\frac{1}{2})^{n-1} + (-\frac{1}{2})^{n-2} + \cdots (-\frac{1}{2})^0 ;\\ = \frac{1 - (-\frac{1}{2})^n}{1 - (-\frac{1}{2})} ;\\ = A ;\\ lim_{n \to \infty} A = \frac{2}{3}