Chapter 12.4, Problem 23E

To determine

To find: The limit of the given function.

Answer to Problem 23E

The limit of the given function is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The given function is,

  $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{5x^3+1}{10x^3-3x^2+7}\right)$

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,

  $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{5x^3+1}{10x^3-3x^2+7}\right)$

The limit of the given functionis calculated as,

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{5x^3+1}{10x^3-3x^2+7}\right)\\ =\underset{x\to -\infty }{\lim }\left(\frac{x^3\left(5+\frac{1}{x^3}\right)}{x^3\left(10-\frac{3}{x}+\frac{7}{x^3}\right)}\right)\\ =\frac{\left(5+\frac{1}{\infty }\right)}{\left(10-\frac{3}{\infty }+\frac{7}{\infty }\right)}\\ =\frac{1}{2}\end{array}$

Consider the graph of the given function as,

  

Therefore, the limit of the given functionis $\frac{1}{2}$ .