Chapter 11.4, Problem 23E

To determine

To Calculate:

The givenlimit. If the limit doesn’t exist then give reason, also check the result graphically.

Answer to Problem 23E

The limit of $f\left(x\right)$ is $\underset{x\to \infty }{\lim }\frac{4t^2+3t-1}{3t^2+2t-5}=\frac{4}{3}$ .

Explanation of Solution

Given Information:

  $\underset{x\to \infty }{\lim }\frac{4t^2+3t-1}{3t^2+2t-5}$

Calculation:

The limit is $\underset{x\to \infty }{\lim }\frac{4t^2+3t-1}{3t^2+2t-5}$

Then,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{4t^2+3t-1}{3t^2+2t-5}=\underset{x\to \infty }{\lim }\frac{4t^2\left(1+\frac{3t}{4t^2}-\frac{1}{4t^2}\right)}{3t^2\left(1+\frac{2t}{3t^2}-\frac{5}{3t^2}\right)}\\ =\underset{x\to \infty }{\lim }\frac{4\left(1+\frac{3t}{4t}-\frac{1}{4t^2}\right)}{3\left(1+\frac{2t}{3t}-\frac{5}{3t^2}\right)}\\ =\frac{4\left(1+\frac{3}{4\cdot \infty }-\frac{1}{4\cdot {\left(\infty \right)}^2}\right)}{3\left(1+\frac{2}{3.\infty }-\frac{5}{3\cdot {\left(\infty \right)}^2}\right)}\\ =\frac{4\left(1+0-0\right)}{3\left(1+0-0\right)}\\ =\frac{4}{3}\end{array}$

Hence the limit of $f\left(x\right)$ is $\underset{x\to \infty }{\lim }\frac{4t^2+3t-1}{3t^2+2t-5}=\frac{4}{3}$ .

To graph the function $f\left(x\right)=\frac{4t^2+3t-1}{3t^2+2t-5}$ ,use the graphing utility

  

The above graph shows the limit function $f\left(x\right)=\frac{4t^2+3t-1}{3t^2+2t-5}$ .