Question

determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
lim_{x
ightarrowinfty}\frac{8x^{2}-10x^{5}}{4 - 3x^{3}+20x^{2}-15x^{5}}

Explanation:

Step1: Divide by highest - power term

Divide both the numerator and denominator by $x^{5}$, since $x^{5}$ is the highest - power term in the denominator.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{8x^{2}-10x^{5}}{4 - 3x^{3}+20x^{2}-15x^{5}}&=\lim_{x ightarrow\infty}\frac{\frac{8x^{2}}{x^{5}}-\frac{10x^{5}}{x^{5}}}{\frac{4}{x^{5}}-\frac{3x^{3}}{x^{5}}+\frac{20x^{2}}{x^{5}}-\frac{15x^{5}}{x^{5}}}\\ &=\lim_{x ightarrow\infty}\frac{\frac{8}{x^{3}} - 10}{\frac{4}{x^{5}}-\frac{3}{x^{2}}+\frac{20}{x^{3}}-15} \end{align*}$$

\]

Step2: Evaluate the limit of each term

As $x
ightarrow\infty$, we know that $\lim_{x
ightarrow\infty}\frac{a}{x^{n}} = 0$ for $a$ being a constant and $n>0$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{\frac{8}{x^{3}} - 10}{\frac{4}{x^{5}}-\frac{3}{x^{2}}+\frac{20}{x^{3}}-15}&=\frac{\lim_{x ightarrow\infty}\frac{8}{x^{3}}-\lim_{x ightarrow\infty}10}{\lim_{x ightarrow\infty}\frac{4}{x^{5}}-\lim_{x ightarrow\infty}\frac{3}{x^{2}}+\lim_{x ightarrow\infty}\frac{20}{x^{3}}-\lim_{x ightarrow\infty}15}\\ &=\frac{0 - 10}{0-0 + 0-15} \end{align*}$$

\]

Step3: Simplify the fraction

\[
\frac{0 - 10}{0-0 + 0-15}=\frac{- 10}{-15}=\frac{2}{3}
\]

Answer:

$\frac{2}{3}$