Chapter 2.1, Problem 26E

To determine

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

Answer to Problem 26E

The value of $\underset{t\to 2}{\lim }\left(\frac{t^2-3t+2}{t^2-4}\right)$ is $0.25$ .

Explanation of Solution

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

Calculation:

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

To graph a function $y=\frac{t^2-3t+2}{t^2-4}$ , 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 $y=\frac{t^2-3t+2}{t^2-4}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{3}\boxed{\text{X}}\boxed{+}\boxed{2}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{4}\boxed{)}$ .

The display will show the equation,

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

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 $\frac{t^2-3t+2}{t^2-4}$ tends to $0.25$ as $t$ tends to $2$ from both the sides. So, the value of $\underset{t\to 2}{\lim }\left(\frac{t^2-3t+2}{t^2-4}\right)$ graphically is $0.25$ .

Check the value of limit algebraically.

  $\begin{array}{c}\underset{t\to 2}{\lim }\left(\frac{t^2-3t+2}{t^2-4}\right)=\underset{t\to 2}{\lim }\left(\frac{t^2-2t-t+2}{\left(t-2\right)\left(t+2\right)}\right)\\ =\underset{t\to 2}{\lim }\left(\frac{t\left(t-2\right)-1\left(t-2\right)}{\left(t-2\right)\left(t+2\right)}\right)\\ =\underset{t\to 2}{\lim }\left(\frac{\left(t-1\right)\left(t-2\right)}{\left(t-2\right)\left(t+2\right)}\right)\\ =\underset{t\to 2}{\lim }\left(\frac{\left(t-1\right)}{\left(t+2\right)}\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{t\to 2}{\lim }\left(\frac{\left(t-1\right)}{\left(t+2\right)}\right)=\left(\frac{2-1}{2+2}\right)\\ =\left(\frac{1}{4}\right)\\ =0.25\end{array}$

So, the value of $\underset{t\to 2}{\lim }\left(\frac{t^2-3t+2}{t^2-4}\right)$ is equal to $0.25$ .

Therefore, the value of $\underset{t\to 2}{\lim }\left(\frac{t^2-3t+2}{t^2-4}\right)$ is $0.25$ .