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} (square) ) (simplify your answer.)

Explanation:

Step1: Factor the denominator

The denominator \(x^2 - 144\) is a difference of squares, so it factors as \((x - 12)(x + 12)\). So we have:
\(\lim_{x
ightarrow 12}\frac{x - 12}{x^2 - 144}=\lim_{x
ightarrow 12}\frac{x - 12}{(x - 12)(x + 12)}\)

Step2: Cancel common factors

We can cancel the common factor of \(x - 12\) (since \(x
ightarrow 12\) but \(x
eq12\) when taking the limit, so we can cancel this factor) to get:
\(\lim_{x
ightarrow 12}\frac{1}{x + 12}\)

Answer:

\(\frac{1}{x + 12}\)