Chapter 11.4, Problem 48E

To determine

To calculate:

The five terms and limit of the sequence, if limit does not exist then give reason.

Answer to Problem 48E

First five terms are $\frac{5}{2},\frac{17}{4},\frac{37}{6},\frac{65}{8},\frac{101}{10}$ and limit of the sequence does not exist.

Explanation of Solution

Given information:

  $a_n=\frac{4n^2+1}{2n}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{4n^2+1}{2n}$

First five terms are : $a_1,a_2,a_3,a_4,a_5$

For $n=1$ :

  $\begin{array}{l}{a}_n=\frac{{4.1}^2+1}{2.1}\\ =\frac{5}{2}\end{array}$

For $n=2$ :

  $\begin{array}{l}{a}_n=\frac{{4.2}^2+1}{2.2}\\ =\frac{17}{4}\end{array}$

For $n=3$ :

  $\begin{array}{l}{a}_n=\frac{{4.3}^2+1}{2.3}\\ =\frac{37}{6}\end{array}$

For $n=4$ :

  $\begin{array}{l}{a}_n=\frac{{4.4}^2+1}{2.4}\\ =\frac{65}{8}\end{array}$

For $n=5$ :

  $\begin{array}{l}{a}_n=\frac{{4.5}^2+1}{2.5}\\ =\frac{101}{10}\end{array}$

Now for finding out the limit of the sequence in case it exists:

For the rational function $f\left(x\right)=\frac{N\left(x\right)}{D\left(x\right)}$

Where,

  $N\left(x\right)=a_n{x}^n+\mathrm{...}+a_0$

  $D\left(x\right)=b_m{x}^m+\mathrm{...}+b_0$

The limit $x$ approaches positive or negative infinity as below:

  $\underset{x\to \infty }{\lim }f\left(x\right)=\{\begin{array}{l}0,ifn<m\\ \frac{a_n}{b_m},ifn=m\end{array}$

The limit does not exist if $n>m$

Hence, the degree of the numerator is grater then the degree of the denominator. So the limit of the function does not exist.