Chapter 2.1, Problem 27E

To determine

To find: The value of $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ with the help of graph and check the results algebraically.

Answer to Problem 27E

The value of $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ is $-0.5$ .

Explanation of Solution

Given information: The limit is $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ .

Calculation:

Let $y=\frac{5x^3+8x^2}{3x^4-16x^2}$ to find the value of $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ graphically.

To graph a function $y=\frac{5x^3+8x^2}{3x^4-16x^2}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{5x^3+8x^2}{3x^4-16x^2}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{5}\boxed{\text{X}}\boxed{^}\boxed{3}\boxed{+}\boxed{8}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}\boxed{÷}\boxed{(}\boxed{3}\boxed{\text{X}}\boxed{^}\boxed{4}\boxed{-}\boxed{16}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(\text{5X}^3+8\text{X}^2\right)/\left(\text{3X}^4-16\text{X}^2\right)$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

The function $\frac{5x^3+8x^2}{3x^4-16x^2}$ tends to $-0.5$ as $x$ tends to $0$ from both the sides. So, the value of $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ graphically is $-0.5$ .

Check the value of limit algebraically.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)=\underset{x\to 0}{\lim }\left(\frac{x^2\left(5x+8\right)}{x^2\left(3x^2-16\right)}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{\left(5x+8\right)}{\left(3x^2-16\right)}\right)\\ =\frac{\left(5\times 0+8\right)}{\left(3\times 0^2-16\right)}\end{array}$

Further simplify.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)=\frac{0+8}{0-16}\\ =\frac{8}{-16}\\ =-0.5\end{array}$

So, the value of $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ is equal to $-0.5$ .

Therefore, the value of $\underset{x\to 0}{\lim }\left(\frac{5x^3+8x^2}{3x^4-16x^2}\right)$ is $-0.5$ .