Question

a. analyze \\(\lim_{x \to \infty} f(x)\\) and \\(\lim_{x \to -\infty} f(x)\\), and then identify the horizontal asymptotes.

\\(\lim_{x \to \infty} \frac{x^4 - 1}{x^2 - 1} = \square\\)

\\(\lim_{x \to -\infty} \frac{x^4 - 1}{x^2 - 1} = \square\\)

identify the horizontal asymptotes. select the correct choice and, if necessary, fill in the answer box(es) to complete your choice.

a. the function has a horizontal asymptote at \\(y = \square\\).
b. the function has two horizontal asymptotes. the top asymptote is \\(y = \square\\) and the bottom asymptote is \\(y = \square\\).
c. the function has no horizontal asymptotes.

b. find the vertical asymptotes. for each vertical asymptote \\(x = a\\), evaluate \\(\lim_{x \to a^-} f(x)\\) and \\(\lim_{x \to a^+} f(x)\\). select the correct choice and, if necessary, fill in the answer boxes to complete your choice.

a. the function has one vertical asymptote at \\(x = \square\\). the limits at this vertical asymptote are \\(\lim_{x \to a^-} f(x) = \square\\) and \\(\lim_{x \to a^+} f(x) = \square\\).
(simplify your answers.)
b. the function has two vertical asymptotes. the leftmost asymptote at \\(x = \square\\) has the limits \\(\lim_{x \to a^-} f(x) = \square\\) and \\(\lim_{x \to a^+} f(x) = \square\\). the rightmost asymptote at \\(x = \square\\) has the limits \\(\lim_{x \to a^-} f(x) = \square\\) and \\(\lim_{x \to a^+} f(x) = \square\\).
(simplify your answers.)
c. the function has no vertical asymptotes.

Explanation:

Evaluate the limits at infinity

$$ \lim_{x \to \infty} \frac{x^4 - 1}{x^2 - 1} = \lim_{x \to \infty} \frac{(x^2 - 1)(x^2 + 1)}{x^2 - 1} = \lim_{x \to \infty} (x^2 + 1) = \infty $$

$$ \lim_{x \to -\infty} \frac{x^4 - 1}{x^2 - 1} = \lim_{x \to -\infty} (x^2 + 1) = \infty $$

Determine horizontal asymptotes

Since both limits as \(x \to \pm\infty\) are infinite, there are no horizontal asymptotes.
The correct multiple-choice option for part (a) is C.

Determine vertical asymptotes

The function is \(f(x) = \frac{x^4 - 1}{x^2 - 1}\).
For \(x
e \pm 1\), \(f(x) = x^2 + 1\).
At \(x = 1\) and \(x = -1\):

$$ \lim_{x \to 1} f(x) = \lim_{x \to 1} (x^2 + 1) = 2 $$

$$ \lim_{x \to -1} f(x) = \lim_{x \to -1} (x^2 + 1) = 2 $$

Since the limits are finite, there are no vertical asymptotes.
The correct multiple-choice option for part (b) is C.

Answer:

Question a

$$ \lim_{x \to \infty} \frac{x^4 - 1}{x^2 - 1} = \infty $$

$$ \lim_{x \to -\infty} \frac{x^4 - 1}{x^2 - 1} = \infty $$

The function has no horizontal asymptotes. (Option C)

Question b

The function has no vertical asymptotes. (Option C)