Chapter 13, Problem 6T

(a)

To determine

To find: The limit of the sequence $a_n=\frac{n}{n^2+4}$.

Answer to Problem 6T

The limit of the sequence is 0.

Explanation of Solution

Given:

The sequence is given,

$a_n=\frac{n}{n^2+4}$

Calculation:

Limit the both sides of the sequence to the infinity,

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{n}{n^2+4}$

Divide the numerator and denominator of rational function by the highest power of n to solve the limit,

$\begin{array}{c}\underset{n\to \infty }{\lim }\frac{n}{n^2+4}=\underset{n\to \infty }{\lim }\frac{\frac{n}{n^2}}{\frac{n^2}{n^2}+\frac{4}{n^2}}\\ =\underset{n\to \infty }{\lim }\frac{\frac{1}{n}}{1+\frac{4}{n^2}}\\ =\frac{\underset{n\to \infty }{\lim }\left(\frac{1}{n}\right)}{\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\left(\frac{4}{n^2}\right)}\end{array}$

Apply limit,

$\begin{array}{c}\underset{n\to \infty }{\lim }\frac{n}{n^2+4}=\frac{\frac{1}{\infty }}{1+\frac{4}{{\infty }^2}}\\ =\frac{0}{1+0}\\ =0\end{array}$

Thus, the limit of the sequence $a_n=\frac{n}{n^2+4}$ is 0.

(b)

To determine

To find: The limit of the sequence $a_n=\sec n\pi$

Answer to Problem 6T

The limit of the sequence $a_n=\sec n\pi$ does not exist.

Explanation of Solution

Given:

The sequence is given,

$a_n=\sec n\pi$

Calculation:

Limit the both sides of the sequence to the infinity,

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\sec n\pi$

For different values of n, $\sec n\pi$ oscillates between $-1$ and 1.

It does not approach any number at infinity.

Thus, the limit of the sequence $a_n=\sec n\pi$ does not exist.