Chapter 2.5, Problem 24E

To determine

To find:the limit of the given function.

Answer to Problem 24E

  $\underset{x\to -\infty }{\lim }\frac{t^2+2}{t^3+t^2-1}=0$ .

Explanation of Solution

Given:

The givenfunction is $\underset{x\to -\infty }{\lim }\frac{t^2+2}{t^3+t^2-1}$ .

Calculation:

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

As $t^2$ grows slower than $t^3$ . The fraction approaches to zero as the denominator increases.

  $\begin{array}{l}\underset{x\to -\infty }{\lim }\frac{t^2+2}{t^3+t^2-1}\left\{\text{given limit}\right\}\\ =\underset{x\to -\infty }{\lim }\frac{2t}{3t^2+2t}\left\{\text{apply L'hospital rule}\right\}\\ =\underset{x\to -\infty }{\lim }\frac{2}{6t+2}\left\{\text{apply again L'hospital rule}\right\}\\ =\underset{x\to -\infty }{\lim }\left(\frac{2}{6\left(-\infty \right)+2}\right)\{\text{plug }x\to \infty \}\\ =\left(\frac{2}{\infty }\right)\left\{\text{any number divided by }\infty \text{=0}\right\}\\ =0\end{array}$

Therefore, $\underset{x\to -\infty }{\lim }\frac{t^2+2}{t^3+t^2-1}=0$