Question

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

a. \lim\limits_{x\to 12} f(x) \qquad b. \lim\limits_{x\to -12^-} f(x) \qquad c. \lim\limits_{x\to -12^+} f(x)

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

now, evaluate the limit select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\bigcirc a. \lim\limits_{x\to 12} f(x)=\square \text{ (simplify your answer.)}
\bigcirc b. the limit does not exist and is neither }\infty \text{ nor } -\infty.

Explanation:

Step1: Simplify the function

We know that \(x^2 - 144=(x - 12)(x + 12)\), so \(\frac{x - 12}{x^2 - 144}=\frac{x - 12}{(x - 12)(x + 12)}=\frac{1}{x + 12}\) (for \(x
eq12\)).

Step2: Evaluate the limit

Now we evaluate \(\lim_{x
ightarrow12}\frac{1}{x + 12}\) using direct substitution. Substitute \(x = 12\) into \(\frac{1}{x + 12}\), we get \(\frac{1}{12+12}=\frac{1}{24}\).

Answer:

A. \(\lim_{x
ightarrow12}f(x)=\frac{1}{24}\)