Question

details
no additional details were added for this assignment.
hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)
score: 1/6 answered: 1/6
question 2
evaluate the limit
lim_{x
ightarrowinfty}\frac{9x + 6}{3x^{2}-4x + 8}

Explanation:

Step1: Divide by highest - power term

Divide both the numerator and denominator by $x^{2}$, the highest - power of $x$ in the denominator. So we have $\lim_{x
ightarrow\infty}\frac{\frac{9x}{x^{2}}+\frac{6}{x^{2}}}{\frac{3x^{2}}{x^{2}}-\frac{4x}{x^{2}}+\frac{8}{x^{2}}}=\lim_{x
ightarrow\infty}\frac{\frac{9}{x}+\frac{6}{x^{2}}}{3-\frac{4}{x}+\frac{8}{x^{2}}}$.

Step2: Use limit properties

We know that $\lim_{x
ightarrow\infty}\frac{c}{x^{n}} = 0$ for any constant $c$ and positive integer $n$. So, $\lim_{x
ightarrow\infty}\frac{9}{x}=0$, $\lim_{x
ightarrow\infty}\frac{6}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{4}{x}=0$, and $\lim_{x
ightarrow\infty}\frac{8}{x^{2}} = 0$.
Then $\lim_{x
ightarrow\infty}\frac{\frac{9}{x}+\frac{6}{x^{2}}}{3-\frac{4}{x}+\frac{8}{x^{2}}}=\frac{0 + 0}{3-0 + 0}$.

Answer:

$0$