Chapter 12, Problem 87RE

To determine

Find the Limit of a Sequence.

Answer to Problem 87RE

Limit of the sequence is $\boxed{\frac{4}{3}}$

Explanation of Solution

Given information:

write the first five terms of the sequence and find the limit of the sequence (if it exists). If the limit does not exist, then explain why. Assume $n$ begins with $n$ .

  $a_n=\frac{4n-1}{3n+1}$

Calculation:

Consider the sequence,

  $a_n=\frac{4n-1}{3n+1},n\ge 1$

Now the first five terms of the sequence is,

  $a_1,a_2,a_3,a_4,a_5.$

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

  $\begin{array}{l}{a}_2=\frac{4n-1}{3n+1}=\frac{4\times 2-1}{3\times 2+1}\\ =\boxed{1}\end{array}$

  $\begin{array}{l}{a}_3=\frac{4n-1}{3n+1}=\frac{4\times 3-1}{3\times 3+1}\\ =\boxed{\frac{11}{10}}\end{array}$

  $\begin{array}{l}{a}_4=\frac{4n-1}{3n+1}=\frac{4\times 4-1}{3\times 4+1}\\ =\boxed{\frac{15}{13}}\end{array}$

  $\begin{array}{l}{a}_5=\frac{4n-1}{3n+1}=\frac{4\times 5-1}{3\times 5+1}\\ =\boxed{\frac{19}{16}}\end{array}$

Find the limit of the sequence, if it exists.

For the rational number,

  $f\left(x\right)=\frac{N\left(x\right)}{D\left(x\right)}$

Where,

  $\begin{array}{l}N\left(x\right)=a_n{x}^n+\mathrm{......}+a_0\\ D\left(x\right)=b_m{x}^m+\mathrm{......}+b_0\end{array}$

The limit as $x$ approaches positive and negative infinity,

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

The limit does not exist if $n>m$ ,

Here the degree of numerator and denominator is equal, so the limit of the function is ratio of the coefficients.

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

Hence limit of the sequence is $\boxed{\frac{4}{3}}$ .