Let g(x)=(-x^(3)+4x)/(4x^(3)-2x^(2)+4). Find lim_(x rarr oo)g(x). Choose 1 answer: (A) -(1)/(4) (B) 1 (C) 0 (D) The limit is unbounded

Given function: We are given the function g(x) = (-x^3 + 4x) / (4x^3 - 2x^2 + 4). To find the limit as x approaches infinity, we can divide the numerator and the denominator by the highest power of x in the denominator, which is x^3.Dividing numerator and denominator: Divide each term in the numerator and the denominator by x^3: g(x) = [(-x^3/x^3) + (4x/x^3)] / [(4x^3/x^3) - (2x^2/x^3) + (4/x^3)]Simplifying terms: Simplify each term: g(x) = [(-1) + (4/x^2)] / [(4) - (2/x) + (4/x^3)]Ignoring terms with x in the denominator: As x approaches infinity, the terms with x in the denominator approach 0. Therefore, we can ignore these terms for the limit calculation: lim_(x -> oo) g(x) = (-1 + 0) / (4 - 0 + 0)Finding the limit: Simplify the expression to find the limit: lim_(x -> oo) g(x) = -1 / 4

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Let
g(x)=(-x^(3)+4x)/(4x^(3)-2x^(2)+4).
Find
lim_(x rarr oo)g(x).
Choose 1 answer:
(A)
-(1)/(4)
(B) 1
(C) 0
(D) The limit is unbounded

Let $g(x)=\frac{-x^{3}+4 x}{4 x^{3}-2 x^{2}+4}$ . $\newline$ Find $\lim _{x \rightarrow \infty} g(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{1}{4}$ $\newline$ (B) $1$ $\newline$ (C) $0$ $\newline$ (D) The limit is unbounded

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Full solution

Q. Let

g(x)=\frac{-x^{3}+4 x}{4 x^{3}-2 x^{2}+4}

\newline

Find

\lim _{x \rightarrow \infty} g(x)

\newline

Choose

1

answer:

\newline

(A)

-\frac{1}{4}

\newline

(B)

1

\newline

(C)

0

\newline

(D) The limit is unbounded

Given function: We are given the function $g(x) = \frac{-x^3 + 4x}{4x^3 - 2x^2 + 4}$ . To find the limit as $x$ approaches infinity, we can divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^3$ .
Dividing numerator and denominator: Divide each term in the numerator and the denominator by $x^3$ : $g(x) = \frac{(-x^3/x^3) + (4x/x^3)}{(4x^3/x^3) - (2x^2/x^3) + (4/x^3)}$
Simplifying terms: Simplify each term: $\newline$ $g(x) = \frac{(-1) + \frac{4}{x^2}}{4 - \frac{2}{x} + \frac{4}{x^3}}$
Ignoring terms with x in the denominator: As $x$ approaches infinity, the terms with $x$ in the denominator approach $0$ . Therefore, we can ignore these terms for the limit calculation: $\lim_{x \to \infty} g(x) = \frac{-1 + 0}{4 - 0 + 0}$
Finding the limit: Simplify the expression to find the limit: $\lim_{x \to \infty} g(x) = -\frac{1}{4}$