Chapter 12.3, Problem 16E

To determine

To find: the limit or state that the limit does not exist and give reason.

Answer to Problem 16E

  $\underset{n\to \infty }{\lim }\frac{6n^2+5}{3n^2}=2.$

Explanation of Solution

Given:

  $\underset{n\to \infty }{\lim }\frac{6n^2+5}{3n^2}$ .

Concept used:

When $x$ becomes infinite, that is when its values become large positive numbers to the extreme right of $0$ .

In that case, $\frac{1}{x}$ becomes very small number, namely $0$ it can be written as:

  $\underset{x\to \infty }{\lim }\frac{1}{x}=0$ .

Whereas when $x$ becomes infinite negatively, that is when its assumes values to the extreme left of $0$ then again $\frac{1}{x}$ approaches $0$ .

  $\underset{x\to -\infty }{\lim }\frac{1}{x}=0$ .

In other words, whenever $x$ becomes infinite positively or negatively, the values of $y=\frac{1}{x}$ approach the horizontal line $y=0$ . That line is called a horizontal asymptote of the grpah.

Calculation:

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{6n^2+5}{3n^2}=\underset{n\to \infty }{\lim }\left(\frac{6n^2}{3n^2}+\frac{5}{3n^2}\right).\\ \Rightarrow \underset{n\to \infty }{\lim }\frac{6n^2}{3n^2}+\underset{n\to \infty }{\lim }\frac{5}{3n^2}.\\ \Rightarrow \frac{6}{3}+\frac{5}{3}\times 0.\\ \Rightarrow 2.\end{array}$

Hence, $\underset{n\to \infty }{\lim }\frac{6n^2+5}{3n^2}=2.$