Question

determine the following limit
lim_{x
ightarrowinfty}\frac{16x^{3}+5x^{2}-3x}{40x^{3}+x^{2}+3x + 4}
select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. (lim_{x
ightarrowinfty}\frac{16x^{3}+5x^{2}-3x}{40x^{3}+x^{2}+3x + 4}=)
(simplify your answer.)
b. the limit as (x) approaches (infty) does not exist and is neither (infty) nor (-infty).

Explanation:

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

When $x\to\infty$, divide both the numerator and denominator of $\frac{16x^{3}+5x^{2}-3x}{40x^{3}+x^{2}+3x + 4}$ by $x^{3}$. We get $\lim_{x\to\infty}\frac{16+\frac{5}{x}-\frac{3}{x^{2}}}{40+\frac{1}{x}+\frac{3}{x^{2}}+\frac{4}{x^{3}}}$.

Step2: Use limit properties

As $x\to\infty$, $\lim_{x\to\infty}\frac{1}{x}=0$, $\lim_{x\to\infty}\frac{1}{x^{2}} = 0$, and $\lim_{x\to\infty}\frac{1}{x^{3}}=0$. So, $\lim_{x\to\infty}\frac{16+\frac{5}{x}-\frac{3}{x^{2}}}{40+\frac{1}{x}+\frac{3}{x^{2}}+\frac{4}{x^{3}}}=\frac{16 + 0-0}{40+0 + 0+0}$.

Step3: Simplify the result

$\frac{16+0 - 0}{40+0+0 + 0}=\frac{16}{40}=\frac{2}{5}$.

Answer:

A. $\frac{2}{5}$