Chapter 12, Problem 24SGA

To determine

To find: the limit and explain the reason if the limit does not exist.

Answer to Problem 24SGA

  $\infty$

Explanation of Solution

Given information:

  $\underset{n\to \infty }{\lim }\frac{2^n{n}^3}{3n^3}$

Calculation:

The limit can be evaluated as follows:

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{2^n{n}^3}{3n^3}=\underset{n\to \infty }{\lim }\left(\frac{1}{n^3}\right)\left(\frac{2^n{n}^3}{3}\right)\\ =\underset{n\to \infty }{\lim }\frac{2^n}{3}\\ =\frac{1}{3}\underset{n\to \infty }{\lim }{2}^n\\ =\frac{1}{3}\underset{n\to \infty }{\lim }{2}^{\infty }\left[\because n=\infty \right]\\ =\left(\frac{1}{3}\right)\left(\infty \right)\left[\underset{n\to \infty }{\lim }{a}^n=\infty ,\text{ if }a>1\right]\\ =\infty \end{array}$

Thus the limit is $\infty$ .