Let h(x)=(-3x^(2)+7)/(x). Find lim_(x rarr oo)h(x). Choose 1 answer: (A) -3 (B) 0 (C) 7 (D) The limit is unbounded

Given function: We are given the function h(x) = (-3x^2 + 7) / x. To find the limit as x approaches infinity, we need to analyze the behavior of the function as x becomes very large.Simplifying the function: We can simplify the function by dividing each term in the numerator by x, which gives us h(x) = -3x + 7/x.Analyzing behavior as x approaches infinity: As x approaches infinity, the term -3x will dominate the behavior of the function because it grows without bound, while the term 7/x will approach 0 because the numerator is constant and the denominator grows without bound.Dominant term in the limit: Therefore, the limit of h(x) as x approaches infinity is dominated by the term -3x, which goes to negative infinity. The term 7/x, which goes to 0, does not affect the limit.Limit of h(x) as x approaches infinity: The limit of h(x) as x approaches infinity is unbounded, and it goes to negative infinity. So the correct answer is (D) The limit is unbounded.

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Let
h(x)=(-3x^(2)+7)/(x).
Find
lim_(x rarr oo)h(x).
Choose 1 answer:
(A) -3
(B) 0
(C) 7
(D) The limit is unbounded

Let $h(x)=\frac{-3 x^{2}+7}{x}$ . $\newline$ Find $\lim _{x \rightarrow \infty} h(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-3$ $\newline$ (B) $0$ $\newline$ (C) $7$ $\newline$ D The limit is unbounded

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Full solution

Q. Let

h(x)=\frac{-3 x^{2}+7}{x}

\newline

Find

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

\newline

Choose

1

answer:

\newline

(A)

-3

\newline

(B)

0

\newline

(C)

7

\newline

D The limit is unbounded

Given function: We are given the function $h(x) = \frac{-3x^2 + 7}{x}$ . To find the limit as $x$ approaches infinity, we need to analyze the behavior of the function as $x$ becomes very large.
Simplifying the function: We can simplify the function by dividing each term in the numerator by $x$ , which gives us $h(x) = -3x + \frac{7}{x}$ .
Analyzing behavior as $x$ approaches infinity: As $x$ approaches infinity, the term $-3x$ will dominate the behavior of the function because it grows without bound, while the term $\frac{7}{x}$ will approach $0$ because the numerator is constant and the denominator grows without bound.
Dominant term in the limit: Therefore, the limit of $h(x)$ as $x$ approaches infinity is dominated by the term $-3x$ , which goes to negative infinity. The term $\frac{7}{x}$ , which goes to $0$ , does not affect the limit.
Limit of $h(x)$ as $x$ approaches infinity: The limit of $h(x)$ as $x$ approaches infinity is unbounded, and it goes to negative infinity. So the correct answer is (D) The limit is unbounded.