Question

Determine ﻿lim⁡x→∞f(x)\$$lim_{x\rightarrow\infty}f$$\left(x\right)limx→∞​f(x)﻿ and ﻿lim⁡x→−∞f(x)\$$lim_{x\rightarrow-\infty}f$$\left(x\right)limx→−∞​f(x)﻿ for the following functions. Then give the horizontal asymptotes of ﻿fff﻿ (if any).f(x)=12x4−4x8−9x4f\left(x\right)=\frac{1}{$$2x^4$$-\sqrt{$$4x^8$-9x^4$$}}

Accepted Answer

Identify the dominant terms in the numerator and the denominator as x approaches infinity and negative infinity. For f(x) = \frac{1}{$$2x^4$$ - \sqrt{$$4x^8$$ - $$9x^4$$}}, the dominant term in the denominator is \sqrt{$$4x^8$$} = $$2x^4$$ when x is large. Simplify the expression by factoring out the dominant term from the square root in the denominator: \sqrt{$$4x^8$$ - $$9x^4$$} = $$x^4$$\sqrt{4 - \frac{9}{$$x^4$$}}. Rewrite the function f(x) in terms of the dominant term: f(x) = \frac{1}{$$2x^4$$ - $$x^4$$\sqrt{4 - \frac{9}{$$x^4$$}}} = \frac{1}{$$x^4(2$$ - \sqrt{4 - \frac{9}{$$x^4$$}})}. Evaluate the limit as x approaches infinity: lim_{x 	o \infty} f(x) = lim_{x 	o \infty} \frac{1}{$$x^4(2$$ - \sqrt{4 - \frac{9}{$$x^4$$}})}. As x becomes very large, \frac{9}{$$x^4$$} approaches 0, so \sqrt{4 - \frac{9}{$$x^4$$}} approaches 2, making the denominator approach $$x^4(2 - 2) = 0. $$Therefore, the limit is undefined, indicating no horizontal asymptote at infinity. Evaluate the limit as x approaches negative infinity: lim_{x 	o -\infty} f(x) = lim_{x 	o -\infty} \frac{1}{$$x^4(2$$ - \sqrt{4 - \frac{9}{$$x^4$$}})}. The analysis is similar to the positive infinity case, and the limit is also undefined, indicating no horizontal asymptote at negative infinity.

Determine $\(\lim\)_{x\(\rightarrow\]\infty\)}f\(\left\)(x\(\right\))$ and $\(\lim\)_{x\(\rightarrow\)-\(\infty\)}f\(\left\)(x\(\right\))$ for the following functions. Then give the horizontal asymptotes of $f$ (if any).

$f (x) = \frac{1}{2 x^{4} - \sqrt{4 x^{8} - 9 x^{4}}}$

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주요 개념

Limits at Infinity

Horizontal Asymptotes

Rational Functions