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).
$f(x)=\dfrac{43x^6 + 3x^2}{19x^5 - 2x}$

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

Explanation:

Step1: Divide numerator and denominator by \(x^5\)

For the function \(f(x)=\frac{43x^6 + 3x^2}{19x^5-2x}\), divide each term in the numerator and denominator by \(x^5\) (the highest power of \(x\) in the denominator). We get:
\(f(x)=\frac{\frac{43x^6}{x^5}+\frac{3x^2}{x^5}}{\frac{19x^5}{x^5}-\frac{2x}{x^5}}=\frac{43x+\frac{3}{x^3}}{19 - \frac{2}{x^4}}\)

Step2: Find \(\lim_{x

ightarrow\infty}f(x)\)
As \(x
ightarrow\infty\), we know that \(\lim_{x
ightarrow\infty}\frac{3}{x^3} = 0\) and \(\lim_{x
ightarrow\infty}\frac{2}{x^4}=0\), and \(\lim_{x
ightarrow\infty}43x=\infty\). So we substitute these limits into the expression for \(f(x)\) after division:
\(\lim_{x
ightarrow\infty}\frac{43x+\frac{3}{x^3}}{19 - \frac{2}{x^4}}=\frac{\lim_{x
ightarrow\infty}(43x+\frac{3}{x^3})}{\lim_{x
ightarrow\infty}(19 - \frac{2}{x^4})}=\frac{\infty+ 0}{19 - 0}=\infty\)

Step3: Find \(\lim_{x

ightarrow-\infty}f(x)\)
As \(x
ightarrow-\infty\), \(\lim_{x
ightarrow-\infty}\frac{3}{x^3}=0\) (since the cube of a negative number is negative and as \(|x|\) becomes large, the fraction approaches 0), \(\lim_{x
ightarrow-\infty}\frac{2}{x^4} = 0\) (even power, so positive and approaches 0), and \(\lim_{x
ightarrow-\infty}43x=-\infty\) (since \(x\) is negative and its magnitude is large). Substitute into the expression:
\(\lim_{x
ightarrow-\infty}\frac{43x+\frac{3}{x^3}}{19 - \frac{2}{x^4}}=\frac{\lim_{x
ightarrow-\infty}(43x+\frac{3}{x^3})}{\lim_{x
ightarrow-\infty}(19 - \frac{2}{x^4})}=\frac{-\infty+0}{19 - 0}=-\infty\)

Step4: Determine horizontal asymptotes

A horizontal asymptote \(y = L\) exists if \(\lim_{x
ightarrow\infty}f(x)=L\) or \(\lim_{x
ightarrow-\infty}f(x)=L\) where \(L\) is a finite number. Since \(\lim_{x
ightarrow\infty}f(x)=\infty\) and \(\lim_{x
ightarrow-\infty}f(x)=-\infty\), there are no horizontal asymptotes.

Answer:

For \(\lim_{x
ightarrow\infty}f(x)\), the limit is \(\infty\) (so option A with the box filled as \(\infty\) is incorrect in the sense of finite limit, but following the calculation, the limit is \(\infty\); for \(\lim_{x
ightarrow-\infty}f(x)\), the limit is \(-\infty\)). And there are no horizontal asymptotes.

(For the first limit \(\lim_{x
ightarrow\infty}f(x)\), the answer is \(\infty\), so if we consider the option A, we fill \(\infty\) in the box. But note that the option B says "the limit does not exist and is neither \(\infty\) nor \(-\infty\)" which is wrong here as the limit is \(\infty\). So the correct choice for \(\lim_{x
ightarrow\infty}f(x)\) is A with \(\lim_{x
ightarrow\infty}f(x)=\infty\))