Question

find the domain of the function f. use limits to describe the behavior of f at value(s) of x not in its domain.
$f(x)=\frac{1}{x+6}$
select the correct choice for the domain below and, if necessary, fill in the answer box to complete your answer.
a. the domain is the set of all real numbers except $x\
eq -6$
(use a comma to separate answers as needed.)
b. the domain is the set of all real numbers.
$\lim\limits_{x\to -6^{-}} f(x)=\square$ (simplify your answer.)
$\lim\limits_{x\to -6^{+}} f(x)=\square$ (simplify your answer.)

Explanation:

Step1: Find Domain

The function is \( f(x)=\frac{1}{x + 6} \). A rational function is undefined when the denominator is zero. So, set \( x+6 = 0 \), which gives \( x=-6 \). Thus, the domain is all real numbers except \( x = -6 \).

Step2: Left - hand Limit (\(x

ightarrow - 6^{-}\))
As \( x \) approaches \( -6 \) from the left (\( x
ightarrow - 6^{-} \)), \( x+6\) approaches \( 0 \) from the negative side (since \( x < - 6\), so \( x + 6<0\)). Then \( \frac{1}{x + 6}\) approaches \(-\infty\) because we have a positive numerator and a negative denominator approaching \(0\). So, \( \lim_{x
ightarrow - 6^{-}}f(x)=-\infty \).

Step3: Right - hand Limit (\(x

ightarrow - 6^{+}\))
As \( x \) approaches \( -6 \) from the right (\( x
ightarrow - 6^{+} \)), \( x + 6\) approaches \( 0 \) from the positive side (since \( x>-6\), so \( x + 6>0\)). Then \( \frac{1}{x + 6}\) approaches \(+\infty\) because we have a positive numerator and a positive denominator approaching \(0\). So, \( \lim_{x
ightarrow - 6^{+}}f(x)=+\infty \).

Answer:

For the domain: A. The domain is the set of all real numbers except \( x
eq - 6 \)
For \( \lim_{x
ightarrow - 6^{-}}f(x) \): \(-\infty\)
For \( \lim_{x
ightarrow - 6^{+}}f(x) \): \(+\infty\)