Question

24,188 Learners found this answer helpful

determine lim f(x) and lim f(x) for the following function. then give the horizontal asymptotes of f, if any.
f(x)=\frac{5x^{3}-7}{x^{4}+3x^{2}}
evaluate lim f(x) select the correct choice below and, if necessary, fill in the answer box to complete your choice
lim_{x \to infty}
oa
lim_{x \to infty}\frac{5x^{3}-7}{x^{4}+3x^{2}}=square \text{ (simplify your answer.)}
ob. the limit does not exist and is neither (infty) nor (-infty).

Explanation:

Step1: Divide numerator and denominator by highest - power of x

Divide both the numerator and denominator of $f(x)=\frac{5x^{3}-7}{x^{4}+3x^{2}}$ by $x^{4}$. We get $\lim_{x
ightarrow\infty}\frac{\frac{5x^{3}}{x^{4}}-\frac{7}{x^{4}}}{\frac{x^{4}}{x^{4}}+\frac{3x^{2}}{x^{4}}}=\lim_{x
ightarrow\infty}\frac{\frac{5}{x}-\frac{7}{x^{4}}}{1 + \frac{3}{x^{2}}}$.

Step2: Evaluate the limit of each term

As $x
ightarrow\infty$, we know that $\lim_{x
ightarrow\infty}\frac{5}{x}=0$, $\lim_{x
ightarrow\infty}\frac{7}{x^{4}} = 0$, and $\lim_{x
ightarrow\infty}\frac{3}{x^{2}}=0$. So, $\lim_{x
ightarrow\infty}\frac{\frac{5}{x}-\frac{7}{x^{4}}}{1+\frac{3}{x^{2}}}=\frac{0 - 0}{1+0}=0$.

Answer:

A. $\lim_{x
ightarrow\infty}\frac{5x^{3}-7}{x^{4}+3x^{2}}=0$