find the limit of the rational function a. as x→∞ and b. as x→ - ∞. write ∞ or - ∞ where appropriate…

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.)

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.)

Answer

Explanation:

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

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

Step2: Evaluate the limit as $x\rightarrow\infty$

As $x\rightarrow\infty$, $\frac{7}{x}\rightarrow0$ and $\frac{20}{x^{2}}\rightarrow0$. So, $\lim_{x\rightarrow\infty}\frac{11}{9+\frac{7}{x}+\frac{20}{x^{2}}}=\frac{11}{9 + 0+0}=\frac{11}{9}$

For $x\rightarrow-\infty$, the steps are the same. Dividing numerator and denominator by $x^{4}$ (since $x^{4}>0$ for all $x\neq0$), we get $\lim_{x\rightarrow-\infty}\frac{11}{9+\frac{7}{x}+\frac{20}{x^{2}}}$. As $x\rightarrow-\infty$, $\frac{7}{x}\rightarrow0$ and $\frac{20}{x^{2}}\rightarrow0$. So $\lim_{x\rightarrow-\infty}\frac{11x^{4}}{9x^{4}+7x^{3}+20x^{2}}=\frac{11}{9}$

Answer:

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