Question

determine $lim_{x
ightarrowinfty}f(x)$ and $lim_{x
ightarrow-infty}f(x)$ for the following function. then give the horizontal asymptotes of $f$, if any.
$f(x)=\frac{5x^{3}-7}{x^{4}+3x^{2}}$
$x
ightarrowinfty$
a. $lim_{x
ightarrowinfty}\frac{5x^{3}-7}{x^{4}+3x^{2}} = 0$ (simplify your answer.)
b. the limit does not exist and is neither $infty$ nor $-infty$.
evaluate $lim_{x
ightarrow-infty}f(x)$. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. $lim_{x
ightarrow-infty}\frac{5x^{3}-7}{x^{4}+3x^{2}}=square$ (simplify your answer.)
b. the limit does not exist and is neither $infty$ nor $-infty$

Explanation:

Step1: Divide by highest - power of x in denominator

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

Step2: Evaluate limits of individual terms

We know that $\lim_{x
ightarrow\pm\infty}\frac{c}{x^{n}} = 0$ for any non - zero constant $c$ and positive integer $n$. So, $\lim_{x
ightarrow\pm\infty}\frac{5}{x}=0$, $\lim_{x
ightarrow\pm\infty}\frac{7}{x^{4}}=0$ and $\lim_{x
ightarrow\pm\infty}\frac{3}{x^{2}}=0$.

Step3: Find the limit

Then $\lim_{x
ightarrow\pm\infty}\frac{\frac{5}{x}-\frac{7}{x^{4}}}{1+\frac{3}{x^{2}}}=\frac{0 - 0}{1+0}=0$.

Answer:

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