Question

consider the following function.
$f(x) = \frac{1}{x^2 - 16}$
complete the following table. (round your answers to two decimal places.)

$x$	$f(x)$
$-4.1$	$\square$
$-4.01$	$\square$
$-4.001$	$\square$
$-3.999$	$\square$
$-3.99$	$\square$
$-3.9$	$\square$
$-3.5$	$\square$

use the table to determine whether $f(x)$ approaches $\infty$ or $-\infty$ as $x$ approaches $-4$ from the left and from the right. use a graphing utility to graph the function to confirm your answer.
$\lim\limits_{x \to -4^-} f(x) = \square$
$\lim\limits_{x \to -4^+} f(x) = \square$

Explanation:

Step1: Calculate f(x) for x < -4

For $x=-4.5$:
$\frac{1}{(-4.5)^2 - 16} = \frac{1}{20.25 - 16} = \frac{1}{4.25} \approx 0.24$
For $x=-4.1$:
$\frac{1}{(-4.1)^2 - 16} = \frac{1}{16.81 - 16} = \frac{1}{0.81} \approx 1.23$
For $x=-4.01$:
$\frac{1}{(-4.01)^2 - 16} = \frac{1}{16.0801 - 16} = \frac{1}{0.0801} \approx 12.48$
For $x=-4.001$:
$\frac{1}{(-4.001)^2 - 16} = \frac{1}{16.008001 - 16} = \frac{1}{0.008001} \approx 124.98$

Step2: Calculate f(x) for x > -4

For $x=-3.999$:
$\frac{1}{(-3.999)^2 - 16} = \frac{1}{15.992001 - 16} = \frac{1}{-0.007999} \approx -125.02$
For $x=-3.99$:
$\frac{1}{(-3.99)^2 - 16} = \frac{1}{15.9201 - 16} = \frac{1}{-0.0799} \approx -12.52$
For $x=-3.9$:
$\frac{1}{(-3.9)^2 - 16} = \frac{1}{15.21 - 16} = \frac{1}{-0.79} \approx -1.27$
For $x=-3.5$:
$\frac{1}{(-3.5)^2 - 16} = \frac{1}{12.25 - 16} = \frac{1}{-3.75} \approx -0.27$

Step3: Find left limit as x→-4⁻

As $x$ approaches -4 from values less than -4, $f(x)$ grows without bound toward $+\infty$.

Step4: Find right limit as x→-4⁺

As $x$ approaches -4 from values greater than -4, $f(x)$ grows without bound toward $-\infty$.

Answer:

Completed Table:

$x$	$f(x)$
-4.1	1.23
-4.01	12.48
-4.001	124.98
-3.999	-125.02
-3.99	-12.52
-3.9	-1.27
-3.5	-0.27

Limits:

$\lim_{x \to -4^-} f(x) = +\infty$
$\lim_{x \to -4^+} f(x) = -\infty$