Question

find each function value and limit. use $-\infty$ or $\infty$ where appropriate.\
$f(x) = \frac{3x^3 - 6x^2}{6x^4 + 3}$\
(a) $f(-6)$\
(b) $f(-12)$\
(c) $\lim\limits_{x \to -\infty} f(x)$\
\
(a) $f(-6) = \square$\
(round to the nearest thousandth as needed.)\
(b) $f(-12) = \square$\
(round to the nearest thousandth as needed.)\
(c) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\
\
\bigcirc a. $\lim\limits_{x \to -\infty} \frac{3x^3 - 6x^2}{6x^4 + 3} = \square$\
(type an integer or a decimal.)\
\bigcirc b. the limit does not exist.

Explanation:

Step1: Calculate \( f(-6) \)

Substitute \( x = -6 \) into \( f(x)=\frac{3x^{3}-6x^{2}}{6x^{4}+3} \).

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LATEXBLOCK0
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Step2: Calculate \( f(-12) \)

Substitute \( x=-12 \) into \( f(x)=\frac{3x^{3}-6x^{2}}{6x^{4}+3} \).

$$
LATEXBLOCK1
$$

Step3: Find \( \lim_{x

ightarrow-\infty}f(x) \)
Divide numerator and denominator by \( x^{4} \) (the highest - power of \( x \) in the denominator).

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LATEXBLOCK2
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As \( x
ightarrow-\infty \), \( \frac{3}{x}
ightarrow0 \), \( \frac{6}{x^{2}}
ightarrow0 \), and \( \frac{3}{x^{4}}
ightarrow0 \).

$$ \lim_{x ightarrow-\infty}\frac{\frac{3}{x}-\frac{6}{x^{2}}}{6+\frac{3}{x^{4}}}=\frac{0 - 0}{6+0}=0 $$

Answer:

(A) \( f(-6)\approx - 0.111 \)
(B) \( f(-12)\approx - 0.049 \)
(C) A. \( \lim_{x
ightarrow-\infty}\frac{3x^{3}-6x^{2}}{6x^{4}+3}=0 \)