Chapter 12.4, Problem 14E

To determine

To evaluate the limit of $\underset{x\to \infty }{\lim }\left(\frac{1-2x}{x+2}\right)$ and verify the result using the graph of $\left(\frac{1-2x}{x+2}\right)$ .

Answer to Problem 14E

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

Explanation of Solution

Given information :

Given function $\underset{x\to \infty }{\lim }\left(\frac{1-2x}{x+2}\right)$ , $Letf(x)=\left(\frac{1-2x}{x+2}\right)$

  $Letf(x)=\left(\frac{1-2x}{x+2}\right)$

  $\begin{array}{l}\underset{x\to \infty }{\lim }\left(\frac{1-2x}{x+2}\right)=\underset{x\to -\infty }{\lim }\left(\frac{x\left(\frac{1}{x}-2\right)}{x\left(1+\frac{2}{x}\right)}\right)\\ \Rightarrow \underset{x\to \infty }{\lim }\left(\frac{1-2x}{x+2}\right)=\underset{x\to -\infty }{\lim }\left(\frac{\left(\frac{1}{x}-2\right)}{\left(1+\frac{2}{x}\right)}\right)\left(\because \underset{x\to \infty }{\lim }\frac{1}{x}=0\right)\\ \end{array}$

Thus, $\underset{x\to \infty }{\lim }\left(\frac{1-2x}{x+2}\right)=-2$ .

Graph: The graph of $f(x)=\left(\frac{3x^2-4}{1-x^2}\right)$ is

  

Interpretation:

From above graph, we observe that the function $f(x)=\left(\frac{1-2x}{x+2}\right)$ is approaches to $-2$ as $x\to \infty$ .Hence, $\underset{x\to \infty }{\lim }\left(\frac{1-2x}{x+2}\right)=-2$