Chapter 11.4, Problem 45E

To determine

To calculate: The first five terms and limit and explain the limits exist or not.

Answer to Problem 45E

The values of terms are $\frac{4}{3},2,\frac{16}{7},\frac{22}{9},\text{ and }\frac{28}{11}$ and limit is 3.

Explanation of Solution

Given information: $a_n=\frac{6n-2}{2n+1}\mathrm{...}\left(1\right)$

To find the required five terms, put $n=1,2,3,4,\text{ and }5$ in equation (1), we get

Calculation:

At$n=1$

  $\begin{array}{l}{a}_1=\frac{6\times 1-2}{2\times 1+1}\\ =\frac{6-2}{2+1}\\ =\frac{4}{3}\end{array}$

At $n=2$

  $\begin{array}{l}{a}_2=\frac{6\times 2-2}{2\times 2+1}\\ =\frac{12-2}{4+1}\\ =\frac{10}{5}\\ =2\end{array}$

At $n=3$

  $\begin{array}{l}{a}_3=\frac{6\times 3-2}{2\times 3+1}\\ =\frac{18-2}{6+1}\\ =\frac{16}{7}\end{array}$

At $n=4$

  $\begin{array}{l}{a}_4=\frac{6\times 4-2}{2\times 4+1}\\ =\frac{24-2}{8+1}\\ =\frac{22}{9}\end{array}$

At $n=5$

  $\begin{array}{l}{a}_1=\frac{6\times 5-2}{2\times 5+1}\\ =\frac{30-2}{10+1}\\ =\frac{28}{11}\end{array}$

Hence, first five terms are $\frac{4}{3},2,\frac{16}{7},\frac{22}{9},\text{ and }\frac{28}{11}$

Now, calculate the limit of equation (1), we get

Let, $\therefore \underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }\frac{6n-2}{2n+1}$

Formula used: $\underset{x\to a}{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)},\left[\because \underset{x\to a}{\lim }g\left(x\right)\ne 0\right]$

  $=\underset{x\to \infty }{\lim }\frac{6n-2}{2n+1}$

Divide by highest denominator power

  $=\underset{n\to \infty }{\lim }\left(\frac{6-\frac{2}{n}}{2+\frac{1}{n}}\right)$

  $=\frac{\underset{n\to \infty }{\lim }\left(6-\frac{2}{n}\right)}{\underset{n\to \infty }{\lim }\left(2+\frac{1}{n}\right)}$

Put, the value of limit

  $\begin{array}{l}=\left(\frac{6-\frac{2}{\infty }}{2+\frac{1}{\infty }}\right)\\ =\frac{6}{2}\\ =3\end{array}$

  $\begin{array}{l}\Rightarrow \underset{x\to {\infty }^{-}}{\lim }\left(\frac{6n-2}{2n+1}\right)+\underset{x\to {\infty }^{+}}{\lim }\left(\frac{6n-2}{2n+1}\right)=3\\ \end{array}$

Therefore, LHL is equal to RHL.

Hence, the limit is exist.