Chapter 11.4, Problem 46E

To determine

To calculate:

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

Answer to Problem 46E

First five terms are $\frac{3}{4},\frac{7}{5},\frac{11}{6},\frac{15}{7},\frac{19}{8}$ and limit of the sequence is $4$ .

Explanation of Solution

Given information:

  $a_n=\frac{4n-1}{n+3}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{4n-1}{n+3}$

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

For $n=1$ :

  $\begin{array}{l}{a}_n=\frac{4\left(1\right)-1}{\left(1\right)+3}\\ =\frac{3}{4}\end{array}$

For $n=2$ :

  $\begin{array}{l}{a}_n=\frac{4\left(2\right)-1}{\left(2\right)+3}\\ =\frac{7}{5}\end{array}$

For $n=3$ :

  $\begin{array}{l}{a}_n=\frac{4\left(3\right)-1}{\left(3\right)+3}\\ =\frac{11}{6}\end{array}$

For $n=4$ :

  $\begin{array}{l}{a}_n=\frac{4\left(4\right)-1}{\left(4\right)+3}\\ =\frac{15}{7}\end{array}$

For $n=5$ :

  $\begin{array}{l}{a}_n=\frac{4\left(5\right)-1}{\left(5\right)+3}\\ =\frac{19}{8}\end{array}$

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 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 sequence is the ratio of the coefficients of the highest powered terms.

  $\begin{array}{l}\underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }\frac{4n-1}{n+3}\\ =\frac{4}{1}\\ =4\end{array}$