Question

find each function value and limit. use -∞ or ∞ where appropriate.
f(x) = \frac{9x^{4}-18x^{2}}{18x^{5}+9}
(a) f(-6)
(b) f(-12)
(c) \lim_{x\to -\infty} f(x)

Explanation:

Step1: Calculate f(-6)

Substitute $x = - 6$ into $f(x)=\frac{9x^{4}-18x^{2}}{18x^{5}+9}$.
$f(-6)=\frac{9\times(-6)^{4}-18\times(-6)^{2}}{18\times(-6)^{5}+9}=\frac{9\times1296 - 18\times36}{18\times(-7776)+9}=\frac{11664-648}{-139968 + 9}=\frac{11016}{-139959}=-\frac{11016}{139959}$

Step2: Calculate f(-12)

Substitute $x=-12$ into $f(x)$.
$f(-12)=\frac{9\times(-12)^{4}-18\times(-12)^{2}}{18\times(-12)^{5}+9}=\frac{9\times20736-18\times144}{18\times(-248832)+9}=\frac{186624 - 2592}{-4478976+9}=\frac{184032}{-4478967}=-\frac{184032}{4478967}$

Step3: Calculate $\lim_{x

ightarrow-\infty}f(x)$
For $\lim_{x
ightarrow-\infty}\frac{9x^{4}-18x^{2}}{18x^{5}+9}$, divide both numerator and denominator by $x^{5}$ (the highest - power of $x$ in the denominator).
We get $\lim_{x
ightarrow-\infty}\frac{\frac{9}{x}-\frac{18}{x^{3}}}{18+\frac{9}{x^{5}}}$.
As $x
ightarrow-\infty$, $\frac{9}{x}
ightarrow0$, $\frac{18}{x^{3}}
ightarrow0$ and $\frac{9}{x^{5}}
ightarrow0$.
So $\lim_{x
ightarrow-\infty}\frac{\frac{9}{x}-\frac{18}{x^{3}}}{18+\frac{9}{x^{5}}}=0$

Answer:

(A) $-\frac{11016}{139959}$
(B) $-\frac{184032}{4478967}$
(C) $0$