Chapter 2.8, Problem 4QR

Solution

To State:

The end behavior of the function $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ using limits.

  $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ approaches $\infty$ at both the left and right ends.

Given:

The function $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ .

Concepts Used:

The end behavior of a function $f\left(x\right)$ is indicated by the values of the limits $\underset{x\to -\infty }{\lim }f\left(x\right)$ and $\underset{x\to \infty }{\lim }f\left(x\right)$ .

  $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\underset{x\to a}{\lim }f\left(x\right)\times \underset{x\to a}{\lim }g\left(x\right)$

  $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}$

  $\underset{x\to a}{\lim }\left(f\left(x\right)\pm g\left(x\right)\right)=\underset{x\to a}{\lim }f\left(x\right)\pm \underset{x\to a}{\lim }g\left(x\right)$

  $\underset{x\to \pm \infty }{\lim }{x}^n\to \left\{\begin{array}{lll}\infty \hfill & ,\hfill & n\text{ is even}\hfill \\ \pm \infty \hfill & ,\hfill & n\text{ is odd}\hfill \end{array}\right.$ if $n>0$ .

  $\underset{x\to \pm \infty }{\lim }\frac{1}{x^n}=0$ if $n>0$ .

Calculations:

Determine the behavior of the function $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ at the left end by calculating $\underset{x\to -\infty }{\lim }f\left(x\right)$ :

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

  $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ approaches $\infty$ at the left end.

Determine the behavior of the function $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ at the right end by calculating $\underset{x\to \infty }{\lim }f\left(x\right)$ :

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

  $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ approaches $\infty$ at the right end.

Conclusion:

  $f\left(x\right)=\frac{2x^3-3x^2+1}{x+1}$ approaches $\infty$ at both the left and right ends.