Question

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

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

b. ( lim_{x \to -12^-} f(x) )

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

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

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

where are the vertical asymptotes of the function, if any exist? select the correct choice below and, if necessary, fill in the answer box to complete your choice

a. the vertical asymptote(s) occur(s) at ( x = square )
(use a comma to separate answers as needed. type an integer or a simplified fraction.)

b. there are no vertical asymptotes

Explanation:

Step1: Factor the denominator

The denominator \(x^2 - 144\) can be factored as a difference of squares: \(x^2 - 144=(x - 12)(x + 12)\). So the function \(f(x)=\frac{x - 12}{(x - 12)(x + 12)}\). We can cancel out the common factor \((x - 12)\) for \(x
eq12\), getting \(f(x)=\frac{1}{x + 12}\) for \(x
eq12\).

Step2: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points, after simplifying). For the simplified function \(y = \frac{1}{x+12}\), the denominator is zero when \(x+12 = 0\), i.e., \(x=- 12\). We also need to check \(x = 12\): when \(x = 12\), the original function has a hole (since the factor \((x - 12)\) cancels), not a vertical asymptote. So the vertical asymptote is at \(x=-12\).

Answer:

The vertical asymptote(s) occur(s) at \(x = - 12\)