Chapter 12, Problem 23SGA

To determine

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

Answer to Problem 23SGA

6

Explanation of Solution

Given information:

  $\underset{n\to \infty }{\lim }\frac{6n-3}{n}$

Calculation:

The limit can be evaluated as follows:

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{6n-3}{n}=\underset{n\to \infty }{\lim }6-\frac{3}{n}\\ =\underset{n\to \infty }{\lim }6-\underset{n\to \infty }{\lim }\frac{3}{n}\\ =\underset{n\to \infty }{\lim }6-\underset{n\to \infty }{\lim }\frac{3}{\infty }\left[\because n=\infty \right]\\ =6-0\\ =6\end{array}$

Thus the limit is 6.