Question

find the limit of the rational function (a) as (x
ightarrowinfty) and (b) as (x
ightarrow-infty). write (infty) or (-infty) where appropriate.
(f(x)=\frac{7x^{7}+5x^{6}+5}{3x^{8}})
(lim_{x
ightarrowinfty}f(x)=square) (simplify your answer.)

Explanation:

Step1: Divide each term by highest - power of x

Divide the numerator and denominator of $f(x)=\frac{7x^{7}+5x^{6}+5}{3x^{8}}$ by $x^{8}$. We get $\frac{\frac{7x^{7}}{x^{8}}+\frac{5x^{6}}{x^{8}}+\frac{5}{x^{8}}}{\frac{3x^{8}}{x^{8}}}=\frac{\frac{7}{x}+\frac{5}{x^{2}}+\frac{5}{x^{8}}}{3}$.

Step2: Find limit as $x\to\infty$

As $x\to\infty$, we know that $\lim_{x\to\infty}\frac{1}{x}=0$, $\lim_{x\to\infty}\frac{1}{x^{2}} = 0$ and $\lim_{x\to\infty}\frac{1}{x^{8}}=0$. So, $\lim_{x\to\infty}\frac{\frac{7}{x}+\frac{5}{x^{2}}+\frac{5}{x^{8}}}{3}=\frac{0 + 0+0}{3}=0$.

Step3: Find limit as $x\to-\infty$

As $x\to-\infty$, $\lim_{x\to-\infty}\frac{1}{x}=0$, $\lim_{x\to-\infty}\frac{1}{x^{2}} = 0$ and $\lim_{x\to-\infty}\frac{1}{x^{8}}=0$. So, $\lim_{x\to-\infty}\frac{\frac{7}{x}+\frac{5}{x^{2}}+\frac{5}{x^{8}}}{3}=\frac{0 + 0+0}{3}=0$.

Answer:

(a) $0$
(b) $0$