Evaluating the Limit of a Rational Function as x Approaches Infinity

Math Problem Statement

Evaluate the limit as $ x \to \infty $ of $ \frac{2x^3 - 4x^2 + 8}{4x^4 + x^2 + 1} $.

The problem is to evaluate the following limit as $x \to \infty$ :

$\lim_{x \to \infty} \frac{2x^3 - 4x^2 + 8}{4x^4 + x^2 + 1}$

To solve this, let's analyze the degrees of the polynomial in the numerator and the denominator:

Since the degree of the denominator is higher than the degree of the numerator, as $x \to \infty$ , the fraction will tend towards zero.

To confirm this rigorously, divide each term in the fraction by $x^4$ (the highest power in the denominator):

= \lim_{x \to \infty} \frac{\frac{2}{x} - \frac{4}{x^2} + \frac{8}{x^4}}{4 + \frac{1}{x^2} + \frac{1}{x^4}}$$ Now, as $$x \to \infty$$, all terms with $$\frac{1}{x}$$, $$\frac{1}{x^2}$$, and $$\frac{1}{x^4}$$ in the numerator and denominator approach zero. Thus, the expression simplifies to: $$\frac{0}{4} = 0$$ ### Final Answer: $$\lim_{x \to \infty} \frac{2x^3 - 4x^2 + 8}{4x^4 + x^2 + 1} = 0$$ Would you like further details or have any questions? --- Here are some related questions to further explore the concept: 1. How would the limit change if the degrees of the numerator and denominator were equal? 2. What if the highest degree term was in the numerator instead? 3. How would this limit change if we were finding $$\lim_{x \to -\infty}$$ instead? 4. Can this type of problem be solved using L'Hôpital's Rule? 5. What role do lower-order terms play in limits as $$x \to \infty$$? **Tip**: When solving limits involving rational functions, comparing the degrees of the polynomials in the numerator and denominator can often provide insight into the limit’s behavior.

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Limits
Polynomials
Asymptotic Behavior

$ \lim_{x \to \infty} \frac{f(x)}{g(x)} $
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