Chapter 2.2, Problem 17E

To determine

To guess: The limit of the function $\underset{x\to 3}{\lim }\frac{x^2-3x}{x^2-9}$.

Answer to Problem 17E

The limit of the function is guessed to 0.5.

Explanation of Solution

Given:

The values of x are $3.1,3.05,3.01,3.001,3.0001,2.9,2.95,2.99,2.999\text{ and }2.9999$.

Calculation:

Since x approaches 3, the values $3.1,3.05,3.01,3.001,\text{ and }3.0001$ are to the right of 3 and the values $2.9,2.95,2.99,2.999\text{ and }2.9999$ are to the left of 3.

Evaluate the function (correct to 6 decimal places) when $x=3.1,3.05,3.01,3.001,\text{ and }3.0001$ as shown in the below table.

$x$	$x^2-3x$	$x^2-9$	$f\left(x\right)=\frac{x^2-3x}{x^2-9}$
3.1	0.31	0.61	0.508197
3.05	0.1525	0.3025	0.504132
3.01	0.0301	0.0601	0.500832
3.001	0.003001	0.006001	0.500083
3.0001	0.0003	0.0006	0.500008

Here, the value of $f\left(x\right)$ gets closer to the value 0.5 as x approaches to 3 from the right side.

That is, $\underset{x\to 3^{+}}{\lim }\frac{x^2-3x}{x^2-9}=0.5$.

Evaluate the function (correct to 6 decimal places) when $x=2.9,2.95,2.99,2.999\text{ and }2.9999$ as shown in the below table.

$x$	$x^2-3x$	$x^2-9$	$f\left(x\right)=\frac{x^2-3x}{x^2-9}$
2.9	−0.29	−0.59	0.491525
2.95	−0.1475	−0.2975	0.495798
2.99	−0.0299	−0.0599	0.499165
2.999	−0.003	−0.006	0.499917
2.9999	−0.0003	−0.0006	0.499992

Here, the value of $f\left(x\right)$ gets closer to the value 0.5 as x approaches to 3 from the left side.

That is, $\underset{x\to 3^{-}}{\lim }\frac{x^2-3x}{x^2-9}=0.5$.

Since the right hand limit and the left hand limits are the same, $\underset{x\to 3}{\lim }\frac{x^2-3x}{x^2-9}$ exists.

Therefore, the limit of the function is guessed to 0.5.