Chapter D, Problem 21E

(a)

To determine

To calculate:

Value of the given limit by using graph.

Answer to Problem 21E

Value of given limit $\underset{n\to \infty }{\lim }\frac{n^5}{n!}$ is $0$ . i.e. $\underset{n\to \infty }{\lim }\frac{n^5}{n!}=0$

Explanation of Solution

Given information:

  $\underset{n\to \infty }{\lim }\frac{n^5}{n!}$

Calculation:

Consider the function,

  $f\left(n\right)=\frac{n^5}{n!}$

Now we will graph the function by pulling the values of $n$ from $1$ to $10$ we have,

  

In the graph we have seen that as $n$ increases, the value of $\frac{n^5}{n!}$ decreases and almost touch to $x-$ axis.

Consider two horizontal lines $y=\epsilon$ and $y=-\epsilon$ such that $\epsilon$ is a very small number,

The value of $\epsilon$ will be small for the large value of $n$ .

Therefore $n\to \infty ,\epsilon \to 0$ .

Hence,

the $\underset{n\to \infty }{\lim }\frac{n^5}{n!}=0$

(b)

To determine

To calculate:

Smallest value of $N$ that correspond to $\epsilon =0.1$ and $\epsilon =0.001$ by using definition $3$ .

Answer to Problem 21E

Smallest value of $N$ will be $12$ .

Explanation of Solution

Given information:

  $\underset{n\to \infty }{\lim }\frac{n^5}{n!}$

Calculation:

If a sequence $\left\{a_n\right\}$ has limit $L$ then,

  $\underset{n\to \infty }{\lim }{a}_n=L$ or $a_n\to L$ as $n\to \infty$

For every $\epsilon >0$ there is corresponding integer $N$ such that

In case $n>N$ then $\left|a_n-L\right|<\epsilon$

Need to find smallest values of $N$ that corresponds to $\epsilon =0.1$ and $\epsilon =0.001$ .

For $\epsilon =0.1$ the graph of the sequence is shown as:

  

In the graph one have seen that for $n=9,\left|\frac{n^3}{n!}\right|>0.1$ but for $n=10,\left|\frac{n^3}{n!}\right|<0.1$ .

Therefore the smallest values of $N$ will be $N=10$ .

For $\epsilon =0.001$ the graph of the sequence is shown as:

  

In the graph one have seen that for $n=11,\left|\frac{n^3}{n!}\right|>0.001$ but for $n=12,\left|\frac{n^3}{n!}\right|<0.001$ .

Therefore, smallest value of $N$ will be $12$ .