Chapter 2.2, Problem 18E

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

$\underset{x\to -3}{\lim }\frac{x^2-3x}{x^2-9},$

x= –2.5, –2.9, –2.95, –2.99, –2.999, –2.9999, –3.5, –3.1, –3.05, –3.01, –3.001, –3.0001

To determine

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

Answer to Problem 18E

The limit of the given function does not exist.

Explanation of Solution

Given:

The values of x are $-2.5,-2.9,-2.95,-2.99,-2.999,-2.9999,-3.5,-3.1,-3.05,-3.01,$

$-3.001\text{ and }-3.0001$.

Calculation:

Since x approaches −3, the values $-3.5,-3.1,-3.05,-3.01,-3.001\text{ and }-3.0001$ are to the left of −3 and the values $-2.5,-2.9,-2.95,-2.99,-2.999,\text{ and}-2.9999$ are to the right of −3.

Evaluate the function (correct to 6 decimal places) when $x=-3.5,-3.1,-3.05,-3.01,-3.001\text{ and }-3.0001$ as shown in the below table.

x	$x^2-3x$	$x^2-9$	$f\left(x\right)=\frac{x^2-3x}{x^2-9}$
−3.5	22.75	3.25	7
−3.1	18.91	0.61	31
−3.05	18.4525	0.3025	61
−3.01	18.0901	0.0601	301
−3.001	18.009	0.006001	3,001
−3.0001	18.0009	0.0006	30,001

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

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

Evaluate the function (correct to 6 decimal places) when $x=-2.5,-2.9,-2.95,-2.99,-2.999,\text{ and}-2.9999$ as shown in the below table.

x	$x^2-3x$	$x^2-9$	$f\left(x\right)=\frac{x^2-3x}{x^2-9}$
−2.5	13.75	−2.75	−5
−2.9	17.11	−0.59	−29
−2.95	17.5525	−0.2975	−59
−2.99	17.9101	−0.0599	−299
−2.999	17.991	−0.006	−2,999
−2.9999	17.9991	−0.0006	−29,999

Here, the value of $f\left(x\right)$ gets closer to the largest positive value as x approaches to 0 from the right side.

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

Since the left-hand limit and the right-hand limit are not the same, the limit of the function does not exist.

Thus, it can be guessed that $\underset{x\to 3}{\lim }\frac{x^2-3x}{x^2-9}$ does not exist.