Let h(x)=(x^(2)+6)/(x^(2)+3). Find lim_(x rarr oo)h(x). Choose 1 answer: (A) 0 (B) 2 (C) 1 (D) The limit is unbounded

Given function: We are given the function h(x) = (x^2 + 6) / (x^2 + 3). To find the limit as x approaches infinity, we can divide the numerator and the denominator by x^2, the highest power of x in the denominator.Dividing numerator and denominator: Divide each term in the numerator and the denominator by x^2: h(x) = (x^2/x^2 + 6/x^2) / (x^2/x^2 + 3/x^2)Simplifying the expression: Simplify the expression by canceling out the x^2 terms and evaluating the limits of the remaining terms as x approaches infinity: h(x) = (1 + 6/x^2) / (1 + 3/x^2)Evaluating the limit: As x approaches infinity, the terms 6/x^2 and 3/x^2 approach 0: lim_(x -> oo) h(x) = (1 + 0) / (1 + 0)Simplifying the limit: Simplify the expression to find the limit: lim_(x -> oo) h(x) = 1 / 1Final value of the limit: The final value of the limit is: lim_(x -> oo) h(x) = 1

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Let
h(x)=(x^(2)+6)/(x^(2)+3).
Find
lim_(x rarr oo)h(x).
Choose 1 answer:
(A) 0
(B) 2
(C) 1
(D) The limit is unbounded

Let $h(x)=\frac{x^{2}+6}{x^{2}+3}$ . $\newline$ Find $\lim _{x \rightarrow \infty} h(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $2$ $\newline$ (C) $1$ $\newline$ (D) The limit is unbounded

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Full solution

Q. Let

h(x)=\frac{x^{2}+6}{x^{2}+3}

\newline

Find

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

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B)

2

\newline

(C)

1

\newline

(D) The limit is unbounded

Given function: We are given the function $h(x) = \frac{x^2 + 6}{x^2 + 3}$ . To find the limit as $x$ approaches infinity, we can divide the numerator and the denominator by $x^2$ , the highest power of $x$ in the denominator.
Dividing numerator and denominator: Divide each term in the numerator and the denominator by $x^2$ : $h(x) = \frac{x^2/x^2 + 6/x^2}{x^2/x^2 + 3/x^2}$
Simplifying the expression: Simplify the expression by canceling out the $x^2$ terms and evaluating the limits of the remaining terms as $x$ approaches infinity: $\newline$ $h(x) = \frac{1 + \frac{6}{x^2}}{1 + \frac{3}{x^2}}$
Evaluating the limit: As $x$ approaches infinity, the terms $\frac{6}{x^2}$ and $\frac{3}{x^2}$ approach $0$ : $\lim_{x \to \infty} h(x) = \frac{(1 + 0)}{(1 + 0)}$
Simplifying the limit: Simplify the expression to find the limit: $\lim_{x \to \infty} h(x) = \frac{1}{1}$
Final value of the limit: The final value of the limit is: $\lim_{x \to \infty} h(x) = 1$