Question

determine \\( \lim\limits_{x \to \infty} f(x) \\) and \\( \lim\limits_{x \to -\infty} f(x) \\) for the following function. then give the horizontal asymptotes of \\( f \\) (if any).\\( f(x) = \frac{43x^6 + 3x^2}{19x^5 - 2x} \\)select the correct choice below and, if necessary, fill in the answer box to complete your choice.\\( \boldsymbol{\checkmark} \\) a. \\( \lim\limits_{x \to \infty} f(x) = \infty \\) (simplify your answer.) b. the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\).select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. \\( \lim\limits_{x \to -\infty} f(x) = \square \\) (simplify your answer.) b. the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\).

Explanation:

Step1: Analyze the leading terms

For \( \lim_{x \to -\infty} \frac{43x^6 + 3x^2}{19x^5 - 2x} \), the leading term of the numerator is \( 43x^6 \) (degree 6, even) and the leading term of the denominator is \( 19x^5 \) (degree 5, odd).

Step2: Simplify the fraction

Divide numerator and denominator by \( x^5 \) (the highest power in the denominator):
\[ \lim_{x \to -\infty} \frac{43x + \frac{3}{x^3}}{19 - \frac{2}{x^4}} \]

Step3: Evaluate the limit

As \( x \to -\infty \), \( \frac{3}{x^3} \to 0 \) and \( \frac{2}{x^4} \to 0 \). The numerator \( 43x \to -\infty \) (since \( x \) is negative and multiplied by positive 43) and the denominator approaches 19. So the limit is \( -\infty \). But wait, let's check the signs again. Wait, \( x^6 \) is positive when \( x \) is negative (even power), \( x^5 \) is negative when \( x \) is negative (odd power). So rewrite the function:
\[ f(x)=\frac{43x^6 + 3x^2}{19x^5 - 2x}=\frac{x^5(43x + \frac{3}{x^3})}{x^5(19 - \frac{2}{x^4})} \quad (x eq0) \]
When \( x \to -\infty \), \( x^5 \) is negative. So the numerator's leading term: \( 43x^6 = 43x \cdot x^5 \), so \( 43x \) when \( x \to -\infty \) is \( -\infty \), times \( x^5 \) (negative) gives positive? Wait, no, let's do it correctly. Let's factor out \( x^6 \) from numerator and \( x^5 \) from denominator:
\[ \frac{43x^6 + 3x^2}{19x^5 - 2x}=\frac{x^6(43 + \frac{3}{x^4})}{x^5(19 - \frac{2}{x^4})}=x \cdot \frac{43 + \frac{3}{x^4}}{19 - \frac{2}{x^4}} \]
Now, as \( x \to -\infty \), \( x \to -\infty \), and \( \frac{43 + \frac{3}{x^4}}{19 - \frac{2}{x^4}} \to \frac{43}{19} \) (positive). So the product is \( -\infty \times \) positive \( = -\infty \)? Wait, no, wait: \( x^6 = x^5 \cdot x \), so when we factor \( x^5 \) from denominator and \( x^6 \) from numerator, we get \( x \times \frac{43 + 3/x^4}{19 - 2/x^4} \). As \( x \to -\infty \), \( x \) is negative, and the fraction \( \frac{43 + 3/x^4}{19 - 2/x^4} \) approaches \( 43/19 \) (positive). So the whole expression approaches \( -\infty \times \) positive \( = -\infty \)? But wait, let's test with a negative number, say \( x = -1000 \):
Numerator: \( 43(-1000)^6 + 3(-1000)^2 = 43\times10^{18} + 3\times10^6 \approx 43\times10^{18} \) (positive)
Denominator: \( 19(-1000)^5 - 2(-1000) = -19\times10^{15} + 2\times10^3 \approx -19\times10^{15} \) (negative)
So positive divided by negative is negative. And the degree of numerator is 6, denominator is 5, so the limit as \( x \to -\infty \) is \( -\infty \) (since numerator degree > denominator degree, and the leading coefficient ratio times \( x^{6 - 5}=x \), and \( x \to -\infty \), so the limit is \( -\infty \)). Wait, but the options: option A is a box, option B is does not exist and neither. Wait, actually, when the limit goes to \( \pm\infty \), the limit does not exist in the real number sense, but in the extended real number system, we say it's \( -\infty \). But the options: option A is "lim f(x) = [box]" (simplify), option B is "the limit does not exist and is neither \( \infty \) nor \( -\infty \)". Wait, no, when the limit is \( \pm\infty \), we say the limit does not exist (in \( \mathbb{R} \)) but we can describe it as \( \pm\infty \). Wait, maybe I made a mistake. Let's re-express the function:

\( f(x) = \frac{43x^6 + 3x^2}{19x^5 - 2x} = \frac{x^5(43x + 3/x^3)}{x^5(19 - 2/x^4)} \) (for \( x eq 0 \)). So cancel \( x^5 \) (note that \( x^5 \) is negative when \( x < 0 \), positive when \( x > 0 \)). So for \( x \to -\infty \), \( x^5 < 0 \), so canceling \( x^5 \) (which is negative) gives:

\( f(x) = \frac{43x + 3/x^3}{19 - 2/x^4} \) but with a sign? Wait, no, actually, \( x^6 = x^5 \cdot x \), so \( \frac{x^6}{x^5} = x \), so the function is \( x \cdot \frac{43 + 3/x^4}{19 - 2/x^4} \). So when \( x \to -\infty \), \( x \to -\infty \), and \( \frac{43 + 3/x^4}{19 - 2/x^4} \to 43/19 \) (positive). So the product is \( -\infty \times \) positive \( = -\infty \). So the limit as \( x \to -\infty \) is \( -\infty \), but does the option A allow that? Wait, the first part (x→∞) was A with ∞, now for x→-∞, let's see:

Wait, maybe I messed up the degree. Numerator degree 6, denominator degree 5. So the limit as \( x \to \pm\infty \) is determined by the leading terms. The leading term of numerator is \( 43x^6 \), denominator is \( 19x^5 \). So \( \frac{43x^6}{19x^5} = \frac{43}{19}x \). So as \( x \to \infty \), \( \frac{43}{19}x \to \infty \), which matches the first part (option A with ∞). As \( x \to -\infty \), \( \frac{43}{19}x \to -\infty \), because \( x \) is negative. So the limit as \( x \to -\infty \) is \( -\infty \). But the options: option A is "lim f(x) = [box]" (simplify), so we can write \( -\infty \), but wait, the problem's second part:

Wait, the second question is about \( \lim_{x \to -\infty} f(x) \). So let's do it again:

\( f(x) = \frac{43x^6 + 3x^2}{19x^5 - 2x} \). Divide numerator and denominator by \( x^5 \) (the highest power in the denominator):

\( \frac{43x + 3/x^3}{19 - 2/x^4} \). Now, as \( x \to -\infty \), \( x \to -\infty \), \( 3/x^3 \to 0 \), \( 2/x^4 \to 0 \). So the numerator approaches \( -\infty \) (since \( x \) is negative and multiplied by 43, positive), and the denominator approaches 19 (positive). So \( \frac{-\infty}{19} = -\infty \). So the limit is \( -\infty \), so option A: \( \lim_{x \to -\infty} f(x) = -\infty \).

Now, horizontal asymptotes: a horizontal asymptote exists if \( \lim_{x \to \infty} f(x) \) or \( \lim_{x \to -\infty} f(x) \) is a finite number. But here, both limits are \( \infty \) and \( -\infty \), so there are no horizontal asymptotes.

But the question first asks for \( \lim_{x \to \infty} \) (already answered as \( \infty \), option A), then \( \lim_{x \to -\infty} \):

Explanation for \( \lim_{x \to -\infty} \):

Step1: Identify leading terms

Numerator: \( 43x^6 \) (degree 6, even), Denominator: \( 19x^5 \) (degree 5, odd).

Step2: Simplify by dividing by \( x^5 \)

\[ \lim_{x \to -\infty} \frac{43x + \frac{3}{x^3}}{19 - \frac{2}{x^4}} \]

Step3: Evaluate limits of terms

As \( x \to -\infty \), \( \frac{3}{x^3} \to 0 \), \( \frac{2}{x^4} \to 0 \), \( 43x \to -\infty \), denominator → 19.

Step4: Determine the limit

The expression becomes \( \frac{-\infty}{19} = -\infty \).

Answer for \( \lim_{x \to -\infty} \):

A. \( \lim_{x \to -\infty} f(x) = -\infty \)

For horizontal asymptotes: Since both \( \lim_{x \to \infty} f(x) = \infty \) and \( \lim_{x \to -\infty} f(x) = -\infty \) (not finite), there are no horizontal asymptotes.

for \( \lim_{x \to -\infty} \):

Step1: Identify leading terms

Numerator: \( 43x^6 \) (degree 6, even), Denominator: \( 19x^5 \) (degree 5, odd).

Step2: Simplify by dividing by \( x^5 \)

\[
\lim_{x \to -\infty} \frac{43x + \frac{3}{x^3}}{19 - \frac{2}{x^4}}
\]

Step3: Evaluate limits of terms

As \( x \to -\infty \), \( \frac{3}{x^3} \to 0 \), \( \frac{2}{x^4} \to 0 \), \( 43x \to -\infty \), denominator → 19.

Step4: Determine the limit

The expression becomes \( \frac{-\infty}{19} = -\infty \).

Answer:

for \( \lim_{x \to -\infty} \):
A. \( \lim_{x \to -\infty} f(x) = -\infty \)

For horizontal asymptotes: Since both \( \lim_{x \to \infty} f(x) = \infty \) and \( \lim_{x \to -\infty} f(x) = -\infty \) (not finite), there are no horizontal asymptotes.