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{2x^{3}+7 + 6x^{4}+21x}{21x^{3}+9x^{5}-28 - 12x^{2}}

Explanation:

Step1: Identify highest - degree terms

For the numerator $2x^{3}+7 + 6x^{4}+21x$, the highest - degree term is $6x^{4}$. For the denominator $21x^{3}+9x^{5}-28 - 12x^{2}$, the highest - degree term is $9x^{5}$.

Step2: Divide numerator and denominator by $x^{5}$

\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{2x^{3}+7 + 6x^{4}+21x}{21x^{3}+9x^{5}-28 - 12x^{2}}&=\lim_{x ightarrow\infty}\frac{\frac{2x^{3}}{x^{5}}+\frac{7}{x^{5}}+\frac{6x^{4}}{x^{5}}+\frac{21x}{x^{5}}}{\frac{21x^{3}}{x^{5}}+\frac{9x^{5}}{x^{5}}-\frac{28}{x^{5}}-\frac{12x^{2}}{x^{5}}}\\ &=\lim_{x ightarrow\infty}\frac{\frac{2}{x^{2}}+\frac{7}{x^{5}}+\frac{6}{x}+\frac{21}{x^{4}}}{\frac{21}{x^{2}} + 9-\frac{28}{x^{5}}-\frac{12}{x^{3}}} \end{align*}$$

\]

Step3: Evaluate the limit

As $x
ightarrow\infty$, $\frac{2}{x^{2}}
ightarrow0$, $\frac{7}{x^{5}}
ightarrow0$, $\frac{6}{x}
ightarrow0$, $\frac{21}{x^{4}}
ightarrow0$, $\frac{21}{x^{2}}
ightarrow0$, $\frac{28}{x^{5}}
ightarrow0$, $\frac{12}{x^{3}}
ightarrow0$.
So, $\lim_{x
ightarrow\infty}\frac{\frac{2}{x^{2}}+\frac{7}{x^{5}}+\frac{6}{x}+\frac{21}{x^{4}}}{\frac{21}{x^{2}} + 9-\frac{28}{x^{5}}-\frac{12}{x^{3}}}=\frac{0 + 0+0 + 0}{0 + 9-0 - 0}=0$

Answer:

$0$