Question

find the limit of the rational function (a) as x→∞ and (b) as x→ -∞. write ∞ or -∞ where appropriate.
f(x) = \frac{7x^{8}+9x^{7}+9}{8x^{9}}
lim f(x) =
x→∞
(simplify your answer.)

Explanation:

Step1: Divide each term by highest - power of x in denominator

Divide each term in the numerator $7x^{8}+9x^{7}+9$ and denominator $8x^{9}$ by $x^{9}$. We get $f(x)=\frac{\frac{7x^{8}}{x^{9}}+\frac{9x^{7}}{x^{9}}+\frac{9}{x^{9}}}{\frac{8x^{9}}{x^{9}}}=\frac{\frac{7}{x}+\frac{9}{x^{2}}+\frac{9}{x^{9}}}{8}$.

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

We know that $\lim_{x\to\infty}\frac{1}{x^{n}} = 0$ for $n>0$. So, $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{\frac{7}{x}+\frac{9}{x^{2}}+\frac{9}{x^{9}}}{8}$.
Using the limit laws for sums and quotients, $\lim_{x\to\infty}\frac{\frac{7}{x}+\frac{9}{x^{2}}+\frac{9}{x^{9}}}{8}=\frac{\lim_{x\to\infty}\frac{7}{x}+\lim_{x\to\infty}\frac{9}{x^{2}}+\lim_{x\to\infty}\frac{9}{x^{9}}}{8}$.
Since $\lim_{x\to\infty}\frac{7}{x}=0$, $\lim_{x\to\infty}\frac{9}{x^{2}} = 0$, and $\lim_{x\to\infty}\frac{9}{x^{9}}=0$, we have $\lim_{x\to\infty}f(x)=\frac{0 + 0+0}{8}=0$.

Answer:

$0$