Question

find the limit of the rational function a. as x→∞ and b. as x→ - ∞. write ∞ or - ∞ where appropriate. h(x)=\frac{11x^{4}}{9x^{4}+7x^{3}+20x^{2}}. a. \lim_{x→∞}\frac{11x^{4}}{9x^{4}+7x^{3}+20x^{2}}=\square (simplify your answer.)

Explanation:

Step1: Divide numerator and denominator by $x^4$

\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{11x^{4}}{9x^{4}+7x^{3}+20x^{2}}&=\lim_{x ightarrow\infty}\frac{\frac{11x^{4}}{x^{4}}}{\frac{9x^{4}}{x^{4}}+\frac{7x^{3}}{x^{4}}+\frac{20x^{2}}{x^{4}}}\\ &=\lim_{x ightarrow\infty}\frac{11}{9 + \frac{7}{x}+\frac{20}{x^{2}}} \end{align*}$$

\]

Step2: Evaluate the limit as $x

ightarrow\infty$
As $x
ightarrow\infty$, $\frac{7}{x}
ightarrow0$ and $\frac{20}{x^{2}}
ightarrow0$. So, $\lim_{x
ightarrow\infty}\frac{11}{9+\frac{7}{x}+\frac{20}{x^{2}}}=\frac{11}{9 + 0+0}=\frac{11}{9}$

For $x
ightarrow-\infty$, the steps are the same. Dividing numerator and denominator by $x^{4}$ (since $x^{4}>0$ for all $x
eq0$), we get $\lim_{x
ightarrow-\infty}\frac{11}{9+\frac{7}{x}+\frac{20}{x^{2}}}$. As $x
ightarrow-\infty$, $\frac{7}{x}
ightarrow0$ and $\frac{20}{x^{2}}
ightarrow0$. So $\lim_{x
ightarrow-\infty}\frac{11x^{4}}{9x^{4}+7x^{3}+20x^{2}}=\frac{11}{9}$

Answer:

a. $\frac{11}{9}$
b. $\frac{11}{9}$