Question

determine \\( \lim_{x \to \infty} f(x) \\) and \\( \lim_{x \to -\infty} f(x) \\) for the following function. then give the horizontal asymptotes of \\( f \\) (if any)\\( f(x) = \frac{4x + 1}{5x^4 + 4} \\)select the correct choice below and, if necessary, fill in the answer box to complete your choice.\\( \boldsymbol{a.} \\) \\( \lim_{x \to \infty} f(x) = \\) (simplify your answer.)\\( \boldsymbol{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.\\( \boldsymbol{a.} \\) \\( \lim_{x \to -\infty} f(x) = \\) (simplify your answer.)\\( \boldsymbol{b.} \\) the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\).identify the horizontal asymptotes. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\\( \boldsymbol{a.} \\) the function has one horizontal asymptote, \\( y = \\)\\( \boldsymbol{b.} \\) the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is (type equations.)\\( \boldsymbol{c.} \\) the function has no horizontal asymptotes.

Explanation:

Step1: Analyze the function for \( x \to \infty \)

For the function \( f(x)=\frac{4x + 1}{5x^4+4} \), as \( x \to \infty \), the highest power of \( x \) in the denominator dominates. We can divide both the numerator and the denominator by \( x^4 \):
\[
\lim_{x \to \infty} \frac{4x + 1}{5x^4+4}=\lim_{x \to \infty} \frac{\frac{4x}{x^4}+\frac{1}{x^4}}{\frac{5x^4}{x^4}+\frac{4}{x^4}}=\lim_{x \to \infty} \frac{\frac{4}{x^3}+\frac{1}{x^4}}{5+\frac{4}{x^4}}
\]
As \( x \to \infty \), \( \frac{4}{x^3} \to 0 \), \( \frac{1}{x^4} \to 0 \), and \( \frac{4}{x^4} \to 0 \). So:
\[
\lim_{x \to \infty} \frac{\frac{4}{x^3}+\frac{1}{x^4}}{5+\frac{4}{x^4}}=\frac{0 + 0}{5+0}=0
\]

Step2: Analyze the function for \( x \to -\infty \)

Similarly, for \( x \to -\infty \), we divide numerator and denominator by \( x^4 \) (note that \( x^4 \) is positive for negative \( x \)):
\[
\lim_{x \to -\infty} \frac{4x + 1}{5x^4+4}=\lim_{x \to -\infty} \frac{\frac{4x}{x^4}+\frac{1}{x^4}}{\frac{5x^4}{x^4}+\frac{4}{x^4}}=\lim_{x \to -\infty} \frac{\frac{4}{x^3}+\frac{1}{x^4}}{5+\frac{4}{x^4}}
\]
As \( x \to -\infty \), \( \frac{4}{x^3} \to 0 \) (since \( x^3 \) is negative and its magnitude increases), \( \frac{1}{x^4} \to 0 \), and \( \frac{4}{x^4} \to 0 \). So:
\[
\lim_{x \to -\infty} \frac{\frac{4}{x^3}+\frac{1}{x^4}}{5+\frac{4}{x^4}}=\frac{0 + 0}{5+0}=0
\]

Step3: Determine horizontal asymptotes

A horizontal asymptote \( y = L \) exists if \( \lim_{x \to \infty} f(x)=L \) or \( \lim_{x \to -\infty} f(x)=L \). Since both \( \lim_{x \to \infty} f(x) = 0 \) and \( \lim_{x \to -\infty} f(x)=0 \), the function has one horizontal asymptote \( y = 0 \).

Answer:

• \( \lim_{x \to \infty} f(x)=\boldsymbol{0} \)
• \( \lim_{x \to -\infty} f(x)=\boldsymbol{0} \)
• The function has one horizontal asymptote, \( y = \boldsymbol{0} \)