Chapter 12.4, Problem 28E

To determine

To find: The limit of the given function.

Answer to Problem 28E

The limit of the given function is $\frac{7}{2}$ .

Explanation of Solution

Given information:

The given function is,

  $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{x}{2x+1}+\frac{3x^2}{{\left(x-3\right)}^2}\right)$

Concept used:

The horizontal asymptote of the function is calculated as,

  $\begin{array}{l}\underset{x\to -\infty }{\lim }f\left(x\right)=L_1\\ \underset{x\to \infty }{\lim }f\left(x\right)=L_2\end{array}$

The given function is,

  $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{x}{2x+1}+\frac{3x^2}{{\left(x-3\right)}^2}\right)$

The limit of the given functionis calculated as,

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{x}{2x+1}+\frac{3x^2}{{\left(x-3\right)}^2}\right)\\ =\underset{x\to -\infty }{\lim }\left(\frac{x}{x\left(2+\frac{1}{x}\right)}+\frac{3x^2}{x^2{\left(1-\frac{3}{x}\right)}^2}\right)\\ =\frac{1}{\left(2+\frac{1}{\infty }\right)}+\frac{3}{{\left(1-\frac{3}{\infty }\right)}^2}\\ =\frac{7}{2}\end{array}$

Consider the graph of the given function as,

  

Therefore, the limit of the given functionis $\frac{7}{2}$ .