Question

find each function value and limit. use -∞ or ∞ where appropriate.
f(x) = \frac{5x^{3}-10x^{2}}{8x^{4}+4}
(a) f(-6)
(b) f(-12)
(c) \lim_{x\to -\infty} f(x)
(a) f(-6)= - 0.139
(round to the nearest thousandth as needed.)
(b) f(-12)=
(round to the nearest thousandth as needed.)

Explanation:

Step1: Substitute x = - 12 into f(x)

$f(-12)=\frac{5(-12)^{3}-10(-12)^{2}}{8(-12)^{4}+4}$

Step2: Calculate the numerator

$5(-12)^{3}-10(-12)^{2}=5\times(-1728)-10\times144=-8640 - 1440=-10080$

Step3: Calculate the denominator

$8(-12)^{4}+4=8\times20736 + 4=165888+4 = 165892$

Step4: Calculate the value of f(-12)

$f(-12)=\frac{-10080}{165892}\approx - 0.061$

Step5: Find the limit as x→ - ∞

Divide both numerator and denominator by $x^{4}$:
$\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}\frac{5x^{3}-10x^{2}}{8x^{4}+4}=\lim_{x
ightarrow-\infty}\frac{\frac{5}{x}-\frac{10}{x^{2}}}{8+\frac{4}{x^{4}}}$
As $x
ightarrow-\infty$, $\frac{5}{x}
ightarrow0$, $\frac{10}{x^{2}}
ightarrow0$ and $\frac{4}{x^{4}}
ightarrow0$
So $\lim_{x
ightarrow-\infty}f(x)=0$

Answer:

(B) - 0.061
(C) 0