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Math  /  Calculus

QuestionMy score: 7.58/107.58 / 10 pts ( 75.83%75.83 \% ) Evaluate lim⁡x→∞f(x)\lim _{x \rightarrow \infty} f(x) and lim⁡x→−∞f(x)\lim _{x \rightarrow-\infty} f(x) for the following function. Then give the horizontal asymptote(s) of ff (if any). f(x)=2x3+84x3+64x6+3f(x)=\frac{2 x^{3}+8}{4 x^{3}+\sqrt{64 x^{6}+3}}
Evaluate lim⁡x→∞f(x)\lim _{x \rightarrow \infty} f(x). Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. lim⁡x→∞f(x)=16\lim _{x \rightarrow \infty} f(x)=\frac{1}{6} (Type an integer or a simplified fraction.) B. The limit does not exist and is neither ∞\infty nor −∞-\infty.
Evaluate lim⁡x→−∞f(x)\lim _{x \rightarrow-\infty} f(x). Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. lim⁡x→−∞f(x)=□\lim _{x \rightarrow-\infty} f(x)=\square (Type an integer or a simplified fraction.) □\square B. The limit does not exist and is neither ∞\infty nor −∞-\infty.

Studdy Solution

STEP 1

1. The function f(x)=2x3+84x3+64x6+3 f(x) = \frac{2x^3 + 8}{4x^3 + \sqrt{64x^6 + 3}} involves a rational expression where both the numerator and the denominator contain polynomial and radical terms.
2. Evaluating the limit as x→∞ x \to \infty and x→−∞ x \to -\infty will require analyzing the highest degree terms in the numerator and the denominator.
3. The horizontal asymptotes of the function will be determined by these limits.

STEP 2

1. Simplify the function f(x) f(x) by factoring out the highest power of x x from both the numerator and the denominator.
2. Evaluate lim⁡x→∞f(x) \lim_{x \to \infty} f(x) .
3. Evaluate lim⁡x→−∞f(x) \lim_{x \to -\infty} f(x) .
4. Determine the horizontal asymptotes based on the limits.

STEP 3

Factor out the highest power of x x in the numerator 2x3+8 2x^3 + 8 .
2x3+8=2x3(1+82x3)=2x3(1+4x3) 2x^3 + 8 = 2x^3 \left( 1 + \frac{8}{2x^3} \right) = 2x^3 \left( 1 + \frac{4}{x^3} \right)

STEP 4

Factor out the highest power of x x in the denominator 4x3+64x6+3 4x^3 + \sqrt{64x^6 + 3} .
64x6+3=64x6(1+364x6)=64x6(1+364x6) 64x^6 + 3 = 64x^6 \left( 1 + \frac{3}{64x^6} \right) = 64x^6 \left( 1 + \frac{3}{64x^6} \right)
4x3+64x6+3=4x3+64x61+364x6=4x3+8x31+364x6 4x^3 + \sqrt{64x^6 + 3} = 4x^3 + \sqrt{64x^6} \sqrt{1 + \frac{3}{64x^6}} = 4x^3 + 8x^3 \sqrt{1 + \frac{3}{64x^6}}

STEP 5

Simplify the expression for f(x) f(x) .
f(x)=2x3(1+4x3)4x3+8x31+364x6=2x3(1+4x3)4x3(1+21+364x6) f(x) = \frac{2x^3 (1 + \frac{4}{x^3})}{4x^3 + 8x^3 \sqrt{1 + \frac{3}{64x^6}}} = \frac{2x^3 (1 + \frac{4}{x^3})}{4x^3 (1 + 2 \sqrt{1 + \frac{3}{64x^6}})}

STEP 6

Evaluate the limit lim⁡x→∞f(x) \lim_{x \to \infty} f(x) .
As x→∞ x \to \infty :
1+4x31+21+364x6≈11+2⋅1=13 \frac{1 + \frac{4}{x^3}}{1 + 2 \sqrt{1 + \frac{3}{64x^6}}} \approx \frac{1}{1 + 2 \cdot 1} = \frac{1}{3}
lim⁡x→∞f(x)=13 \lim_{x \to \infty} f(x) = \frac{1}{3}

STEP 7

Evaluate the limit lim⁡x→−∞f(x) \lim_{x \to -\infty} f(x) .
As x→−∞ x \to -\infty :
1+4x31+21+364x6≈11+2⋅1=13 \frac{1 + \frac{4}{x^3}}{1 + 2 \sqrt{1 + \frac{3}{64x^6}}} \approx \frac{1}{1 + 2 \cdot 1} = \frac{1}{3}
lim⁡x→−∞f(x)=13 \lim_{x \to -\infty} f(x) = \frac{1}{3}

STEP 8

Determine the horizontal asymptotes based on the limits.
Since both lim⁡x→∞f(x) \lim_{x \to \infty} f(x) and lim⁡x→−∞f(x) \lim_{x \to -\infty} f(x) are equal to 13 \frac{1}{3} , the horizontal asymptote is:
y=13 y = \frac{1}{3}
Solution: - The limit as x→∞ x \to \infty is 13 \frac{1}{3} . - The limit as x→−∞ x \to -\infty is 13 \frac{1}{3} . - The horizontal asymptote of f f is y=13 y = \frac{1}{3} .

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