Question

analyze the following limits and find the vertical asymptotes of ( f(x) = \frac{x - 12}{x^2 - 144} ).

a. ( limlimits_{x \to 12} f(x) )

b. ( limlimits_{x \to -12^-} f(x) )
c. ( limlimits_{x \to -12^+} f(x) )

a. ( limlimits_{x \to 12} f(x) = \frac{1}{24} ) (simplify your answer.)

b. the limit does not exist and is neither ( infty ) nor ( -infty ).

b. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. ( limlimits_{x \to -12^-} f(x) = -infty ) (simplify your answer.)

b. the limit does not exist and is neither ( infty ) nor ( -infty ).

c. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. ( limlimits_{x \to -12^+} f(x) = square ) (simplify your answer.)

b. the limit does not exist and is neither ( infty ) nor ( -infty ).

Explanation:

Step1: Simplify the function

First, factor the denominator: \(x^2 - 144=(x - 12)(x + 12)\). So the function \(f(x)=\frac{x - 12}{(x - 12)(x + 12)}\), and we can cancel \(x - 12\) (for \(x
eq12\)), getting \(f(x)=\frac{1}{x + 12}\) (for \(x
eq12\)).

Step2: Analyze \(\lim_{x

ightarrow - 12^{+}}f(x)\)
As \(x
ightarrow - 12^{+}\), \(x+12
ightarrow0^{+}\) (since \(x\) is approaching \(- 12\) from the right, \(x+12\) is a small positive number). Then \(\frac{1}{x + 12}\) will approach \(+\infty\) because the numerator is \(1\) (positive) and the denominator approaches \(0\) from the positive side.

Answer:

A. \(\lim\limits_{x
ightarrow - 12^{+}}f(x)=\infty\) (or more precisely \(+\infty\))