Chapter 12, Problem 25SGA

To determine

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

Answer to Problem 25SGA

0

Explanation of Solution

Given information:

  $\underset{n\to \infty }{\lim }\frac{4n^3-3n}{n^4-4n^3}$

Calculation:

The limit can be evaluated as follows:

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{4n^3-3n}{n^4-4n^3}=\underset{n\to \infty }{\lim }\frac{\frac{4n^3-3n}{n^4}}{\frac{n^4-4n^3}{n^4}}\\ =\underset{n\to \infty }{\lim }\frac{\frac{4}{n}-\frac{3}{n^3}}{1-\frac{4}{n}}\\ =\frac{\underset{n\to \infty }{\lim }\frac{4}{n}-\underset{n\to \infty }{\lim }\frac{3}{n^3}}{1-\underset{n\to \infty }{\lim }\frac{4}{n}}\\ =\frac{0-0}{1-0}\left[\because n=\infty \right]\\ =\frac{0}{1}\\ =0\end{array}$

Thus the limit is 0.