Chapter 2.2, Problem 50E

To determine

To find: The value of $\underset{x\to \infty }{\lim }f\left(x\right)$ and $\underset{x\to -\infty }{\lim }f\left(x\right)$ for $f\left(x\right)=x^2{e}^{-x}$ with the help of graph of $y=f\left(\frac{1}{x}\right)$ .

Answer to Problem 50E

The value of $\underset{x\to \infty }{\lim }f\left(x\right)$ and $\underset{x\to -\infty }{\lim }f\left(x\right)$ for $f\left(x\right)=x^2{e}^{-x}$ are $0$ and $\infty$ respectively.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=x^2{e}^{-x}$ .

Calculation:

The value of $\underset{x\to \infty }{\lim }f\left(x\right)$ is equal to the value of $\underset{x\to 0^{+}}{\lim }\left(f\left(\frac{1}{x}\right)\right)$ and the value of $\underset{x\to -\infty }{\lim }f\left(x\right)$ is equal to the value of $\underset{x\to 0^{-}}{\lim }\left(f\left(\frac{1}{x}\right)\right)$ .

Simplify the function.

  $\begin{array}{c}y=f\left(\frac{1}{x}\right)\\ ={\left(\frac{1}{x}\right)}^2{e}^{-\frac{1}{x}}\\ =\frac{e^{-\frac{1}{x}}}{x^2}\end{array}$

To graph a function $y=\frac{e^{-\frac{1}{x}}}{x^2}$ , 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 $y=\frac{e^{-\frac{1}{x}}}{x^2}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{2\text{nd}}\boxed{\text{LN}}\boxed{(-)}\boxed{1}\boxed{÷}\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(e^\left(-1/\text{X}\right)\right)/\left(\text{X}^2\right)$

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)

As observed from graph the value of $y=\frac{e^{-\frac{1}{x}}}{x^2}$ approaches to $0$ as $x$ approaches to $0$ from right side.So, the value of $\underset{x\to \infty }{\lim }\left(x^2{e}^{-x}\right)=0$ .

It is also observed from the graph the value of $y=\frac{e^{-\frac{1}{x}}}{x^2}$ approaches to $\infty$ as $x$ approaches to $0$ from left side.So, the value of $\underset{x\to -\infty }{\lim }\left(x^2{e}^{-x}\right)=\infty$ .

Therefore, the value of $\underset{x\to \infty }{\lim }f\left(x\right)$ and $\underset{x\to -\infty }{\lim }f\left(x\right)$ for $f\left(x\right)=x^2{e}^{-x}$ are $0$ and $\infty$ respectively.