Let g(x)=(5x)/(5x^(4)-4). Find lim_(x rarr oo)g(x). Choose 1 answer: (A) -(5)/(4) (B) 1 (C) 0 (D) The limit is unbounded

Given function: We are given the function g(x) = (5x) / (5x^4 - 4). To find the limit as x approaches infinity, we can analyze the behavior of the numerator and the denominator separately.Numerator behavior: As x approaches infinity, the term 5x in the numerator also approaches infinity. However, the rate at which the numerator grows is much slower compared to the denominator.Denominator behavior: In the denominator, the term 5x^4 dominates because it grows much faster than the constant -4 as x approaches infinity. Therefore, the -4 becomes negligible in comparison to 5x^4.Simplifying the expression: We can divide both the numerator and the denominator by x^4, the highest power of x in the denominator, to simplify the expression. g(x) = (5x) / (5x^4 - 4) = (5x/x^4) / (5x^4/x^4 - 4/x^4)Limit as x approaches infinity: Simplifying the expression, we get: g(x) = (5/x^3) / (5 - 4/x^4) As x approaches infinity, 5/x^3 approaches 0 and 4/x^4 also approaches 0.The simplified expression becomes: g(x) = 0 / (5 - 0) = 0 / 5Therefore, the limit of g(x) as x approaches infinity is 0.

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Let
g(x)=(5x)/(5x^(4)-4).
Find
lim_(x rarr oo)g(x).
Choose 1 answer:
(A)
-(5)/(4)
(B) 1
(C) 0
(D) The limit is unbounded

Let $g(x)=\frac{5 x}{5 x^{4}-4}$ . $\newline$ Find $\lim _{x \rightarrow \infty} g(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{5}{4}$ $\newline$ (B) $1$ $\newline$ (C) $0$ $\newline$ (D) The limit is unbounded

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Full solution

Q. Let

g(x)=\frac{5 x}{5 x^{4}-4}

\newline

Find

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

\newline

Choose

1

answer:

\newline

(A)

-\frac{5}{4}

\newline

(B)

1

\newline

(C)

0

\newline

(D) The limit is unbounded

Given function: We are given the function $g(x) = \frac{5x}{5x^4 - 4}$ . To find the limit as $x$ approaches infinity, we can analyze the behavior of the numerator and the denominator separately.
Numerator behavior: As $x$ approaches infinity, the term $5x$ in the numerator also approaches infinity. However, the rate at which the numerator grows is much slower compared to the denominator.
Denominator behavior: In the denominator, the term $5x^4$ dominates because it grows much faster than the constant $-4$ as $x$ approaches infinity. Therefore, the $-4$ becomes negligible in comparison to $5x^4$ .
Simplifying the expression: We can divide both the numerator and the denominator by $x^4$ , the highest power of $x$ in the denominator, to simplify the expression. $\newline$ $g(x) = \frac{5x}{5x^4 - 4} = \frac{\frac{5x}{x^4}}{\frac{5x^4}{x^4} - \frac{4}{x^4}}$
Limit as $x$ approaches infinity: Simplifying the expression, we get: $g(x) = \frac{5/x^3}{5 - 4/x^4}$ As $x$ approaches infinity, $5/x^3$ approaches $0$ and $4/x^4$ also approaches $0$ .
Limit as $x$ approaches infinity: Simplifying the expression, we get: $g(x) = \frac{5/x^3}{5 - 4/x^4}$ As $x$ approaches infinity, $5/x^3$ approaches $0$ and $4/x^4$ also approaches $0$ .The simplified expression becomes: $g(x) = \frac{0}{5 - 0} = \frac{0}{5}$
Limit as $x$ approaches infinity: Simplifying the expression, we get: $g(x) = \frac{5/x^3}{5 - 4/x^4}$ As $x$ approaches infinity, $5/x^3$ approaches $0$ and $4/x^4$ also approaches $0$ .The simplified expression becomes: $g(x) = \frac{0}{5 - 0} = \frac{0}{5}$ Therefore, the limit of $g(x)$ as $x$ approaches infinity is $0$ .