数学 - 無限の世界への一歩 - 数列の極限、無限級数 – 極限の計算 - 極限の法則(1) - 極限値と四則

数学読本〈4〉

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数学読本〈4〉数列の極限，無限級数/順列・組合せ/確率/関数の極限と微分法(松坂 和夫(著)、岩波書店)の第14章(無限の世界への一歩 - 数列の極限、無限級数)、14.2(極限の計算)、極限の法則(1) - 極限値と四則、問1.を取り組んでみる。

2. 1. 
      $2$
   2. 
      $\frac{2}{5}$
   3. 
      $\underset{n\to \infty }{\lim }\frac{\frac{5}{n^2}-3}{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}=-\frac{3}{2}$
   4. 
      $0$
3. 1. 
      $\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n+1}+\sqrt{n}}=0$
   2. 
      $\underset{n\to \infty }{\lim }\frac{-2n}{\sqrt{n^2-2n}+n}=\underset{n\to \infty }{\lim }\frac{-2}{\sqrt{1-\frac{2}{n}}+1}=-1$
   3. 
      $\underset{n\to \infty }{\lim }\frac{2n}{\sqrt{n^2+n+1}+\sqrt{n^2-n+1}}=\frac{2}{1+1}=1$
   4. 
      $\underset{n\to \infty }{\lim }\frac{\sqrt{4n^2+n}+2n}{n}=\sqrt{4}+2=4$
4. 
   $\underset{n\to \infty }{\lim }\frac{\left(n+2\right)\left(n+3\right)}{n\left(n+1\right)}=\frac{1}{1}=1$
5. 1. 
      $\underset{n\to \infty }{\lim }\frac{n\left(1+n\right)}{2n^2}=\frac{1}{2}$
   2. 
      $\begin{array}{l}{(x+1)}^3=x^3+3x^2+3x+1\hfill \\ {(x+1)}^3-x^3=3x^2+3x+1\hfill \\ {\displaystyle \sum }_{x=1}^n({(x+1)}^3-x^3)={(n+1)}^3-1\hfill \\ {\displaystyle \sum }_{x=1}^n(3x^2+3x+1)=3{\displaystyle \sum }_{x=1}^n{x}^2+3{\displaystyle \sum }_{x=1}^{n}x+n\hfill \\ {(n+1)}^3-1=3{\displaystyle \sum }_{x=1}^n{x}^2+\frac{3n(n+1)}{2}+n\hfill \\ {\displaystyle \sum }_{x=1}^n{x}^2=\frac{1}{3}({(n+1)}^3-1-\frac{3n(n+1)}{2}-n)\hfill \\ =\frac{2n^3+6n^2+6n+2-2-3n^2-3n-2n}{6}\hfill \\ =\frac{2n^3+3n^2+n}{6}\hfill \\ =\frac{n(n+1)(2n+1)}{6}\hfill \\ \underset{n\to \infty }{\lim }\frac{1}{n^3}·\frac{n(n+1)(2n+1)}{6}=\frac{2}{6}=\frac{1}{3}\hfill \end{array}$
   3. 
      $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{1}{n^3}\left(2·\frac{n\left(2n+1\right)\left(n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{1·\left(2+\frac{1}{n}\right)·\left(1+\frac{1}{n}\right)}{3}\right)+0\\ =\frac{2}{3}\end{array}$
6. 1. 
      $0$
   2. 
      $1$
   3. 
      $\underset{n\to \infty }{\lim }{\log }_2\frac{\sqrt{n}}{\sqrt{16n+5}}=\underset{n\to \infty }{\lim }{\log }_2\frac{1}{\sqrt{16+\frac{5}{n}}}={\log }_2\frac{1}{4}=-2$
7. 
   $\begin{array}{l}\underset{n\to \infty }{\lim }\left(\frac{1}{n}·\frac{b^2+c^2}{{\left(n+1\right)}^2}\frac{n\left(2n+1\right)\left(n+1\right)}{6}\right)\\ =\underset{n\to \infty }{\lim }\frac{a^{2}n\left(2n+1\right)\left(n+1\right)}{6n{\left(n+1\right)}^2}\\ =\frac{a^2·2}{6}\\ =\frac{a^2}{3}\end{array}$