Chapter 11.4, Problem 18E

To determine

To calculate: The value of limit and explain the limits exist or not.

Answer to Problem 18E

The value of limit is 2.

Explanation of Solution

Given information: $\underset{x\to -\infty }{\lim }\left(\frac{8x+7}{4x+9}\right)$

Formula used: $\underset{x\to a}{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)},\left[\because \underset{x\to a}{\lim }g\left(x\right)\ne 0\right]$

Calculation:

Divide by highest denominator power:

  $\begin{array}{l}=\underset{x\to -\infty }{\lim }\left(\frac{8+\frac{7}{x}}{4+\frac{9}{x}}\right)\\ =\frac{\underset{x\to -\infty }{\lim }\left(8+\frac{7}{x}\right)}{\underset{x\to -\infty }{\lim }\left(4+\frac{9}{x}\right)}\end{array}$

Put, the value of limit

  $=\frac{\left(8+\frac{7}{\infty }\right)}{\left(4+\frac{9}{\infty }\right)}$

  $\begin{array}{l}=\frac{8}{4}\\ =2\end{array}$

  $\therefore \underset{x\to -\infty }{\lim }\left(\frac{8x+7}{4x+9}\right)=2$

  $\Rightarrow \underset{x\to -{\infty }^{-}}{\lim }\left(\frac{8x+7}{4x+9}\right)=\underset{x\to -{\infty }^{+}}{\lim }\left(\frac{8x+7}{4x+9}\right)=2$

Therefore, LHL is equal to RHL.

Hence, the limit is exist.