## 2017年2月3日金曜日

### 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

1. $2$

2. $\frac{2}{5}$

3. $\underset{n\to \infty }{\mathrm{lim}}\frac{\frac{5}{{n}^{2}}-3}{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}=-\frac{3}{2}$

4. $0$

1. $\underset{n\to \infty }{\mathrm{lim}}\frac{1}{\sqrt{n+1}+\sqrt{n}}=0$

2. $\underset{n\to \infty }{\mathrm{lim}}\frac{-2n}{\sqrt{{n}^{2}-2n}+n}=\underset{n\to \infty }{\mathrm{lim}}\frac{-2}{\sqrt{1-\frac{2}{n}}+1}=-1$

3. $\underset{n\to \infty }{\mathrm{lim}}\frac{2n}{\sqrt{{n}^{2}+n+1}+\sqrt{{n}^{2}-n+1}}=\frac{2}{1+1}=1$

4. $\underset{n\to \infty }{\mathrm{lim}}\frac{\sqrt{4{n}^{2}+n}+2n}{n}=\sqrt{4}+2=4$

1. $\underset{n\to \infty }{\mathrm{lim}}\frac{\left(n+2\right)\left(n+3\right)}{n\left(n+1\right)}=\frac{1}{1}=1$

1. $\underset{n\to \infty }{\mathrm{lim}}\frac{n\left(1+n\right)}{2{n}^{2}}=\frac{1}{2}$

2. $\begin{array}{l}{\left(x+1\right)}^{3}={x}^{3}+3{x}^{2}+3x+1\hfill \\ {\left(x+1\right)}^{3}-{x}^{3}=3{x}^{2}+3x+1\hfill \\ \sum _{x=1}^{n}\left({\left(x+1\right)}^{3}-{x}^{3}\right)={\left(n+1\right)}^{3}-1\hfill \\ \sum _{x=1}^{n}\left(3{x}^{2}+3x+1\right)=3\sum _{x=1}^{n}{x}^{2}+3\sum _{x=1}^{n}x+n\hfill \\ {\left(n+1\right)}^{3}-1=3\sum _{x=1}^{n}{x}^{2}+\frac{3n\left(n+1\right)}{2}+n\hfill \\ \sum _{x=1}^{n}{x}^{2}=\frac{1}{3}\left({\left(n+1\right)}^{3}-1-\frac{3n\left(n+1\right)}{2}-n\right)\hfill \\ =\frac{2{n}^{3}+6{n}^{2}+6n+2-2-3{n}^{2}-3n-2n}{6}\hfill \\ =\frac{2{n}^{3}+3{n}^{2}+n}{6}\hfill \\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}\hfill \\ \underset{n\to \infty }{\mathrm{lim}}\frac{1}{{n}^{3}}·\frac{n\left(n+1\right)\left(2n+1\right)}{6}=\frac{2}{6}=\frac{1}{3}\hfill \end{array}$

3. $\begin{array}{l}\underset{n\to \infty }{\mathrm{lim}}\frac{1}{{n}^{3}}\left(2·\frac{n\left(2n+1\right)\left(n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right)\\ =\underset{n\to \infty }{\mathrm{lim}}\left(\frac{1·\left(2+\frac{1}{n}\right)·\left(1+\frac{1}{n}\right)}{3}\right)+0\\ =\frac{2}{3}\end{array}$

1. $0$

2. $1$

3. $\underset{n\to \infty }{\mathrm{lim}}{\mathrm{log}}_{2}\frac{\sqrt{n}}{\sqrt{16n+5}}=\underset{n\to \infty }{\mathrm{lim}}{\mathrm{log}}_{2}\frac{1}{\sqrt{16+\frac{5}{n}}}={\mathrm{log}}_{2}\frac{1}{4}=-2$

2. $\begin{array}{l}\underset{n\to \infty }{\mathrm{lim}}\left(\frac{1}{n}·\frac{{b}^{2}+{c}^{2}}{{\left(n+1\right)}^{2}}\frac{n\left(2n+1\right)\left(n+1\right)}{6}\right)\\ =\underset{n\to \infty }{\mathrm{lim}}\frac{{a}^{2}n\left(2n+1\right)\left(n+1\right)}{6n{\left(n+1\right)}^{2}}\\ =\frac{{a}^{2}·2}{6}\\ =\frac{{a}^{2}}{3}\end{array}$