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. rewrite the given limit in a way that can be evaluated using direct substitution
( lim_{x \to 12} \frac{x - 12}{x^2 - 144} = lim_{x \to 12} left( \frac{1}{x + 12}
ight) ) (simplify your answer.)

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

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

Explanation:

Part b:

Step1: Recall the simplified function

We know \( f(x)=\frac{x - 12}{x^2-144}=\frac{1}{x + 12} \) (for \( x
eq12 \) and \( x
eq - 12 \)).

Step2: Analyze the limit as \( x

ightarrow - 12^{-} \)
As \( x
ightarrow - 12^{-} \), \( x+12
ightarrow0^{-} \) (since \( x \) is slightly less than - 12, so \( x + 12\) is a small negative number). Then \( \frac{1}{x + 12}
ightarrow-\infty \) because we are dividing 1 by a small negative number.

Answer:

\(\lim\limits_{x
ightarrow - 12^{-}}f(x)=-\infty\)