To solve the given limit:

$\lim_{x \to \infty} \frac{4x^4 + 3x^2 + 4x + 5}{2x^4 + 3x^2 + 4}$

Step-by-step solution:

1. Identify the highest degree of $x$:

   • The highest degree of $x$ in the numerator is $x^4$.
   • The highest degree of $x$ in the denominator is also $x^4$.

2. Divide both the numerator and denominator by $x^4$: $\lim_{x \to \infty} \frac{\frac{4x^4 + 3x^2 + 4x + 5}{x^4}}{\frac{2x^4 + 3x^2 + 4}{x^4}}$

   Simplify each term:

   • Numerator: $\frac{4x^4}{x^4} + \frac{3x^2}{x^4} + \frac{4x}{x^4} + \frac{5}{x^4} = 4 + \frac{3}{x^2} + \frac{4}{x^3} + \frac{5}{x^4}$
   • Denominator: $\frac{2x^4}{x^4} + \frac{3x^2}{x^4} + \frac{4}{x^4} = 2 + \frac{3}{x^2} + \frac{4}{x^4}$

3. Substitute the simplified forms into the limit: $\lim_{x \to \infty} \frac{4 + \frac{3}{x^2} + \frac{4}{x^3} + \frac{5}{x^4}}{2 + \frac{3}{x^2} + \frac{4}{x^4}}$

4. Evaluate the limit as $x \to \infty$:

   • As $x \to \infty$, all terms containing $x$ in the denominator ($\frac{3}{x^2}, \frac{4}{x^3}, \frac{5}{x^4}$) approach $0$.
   • The numerator approaches $4$, and the denominator approaches $2$.

   Therefore: $\lim_{x \to \infty} \frac{4 + \frac{3}{x^2} + \frac{4}{x^3} + \frac{5}{x^4}}{2 + \frac{3}{x^2} + \frac{4}{x^4}} = \frac{4}{2} = 2$

Final Answer:

$\lim_{x \to \infty} \frac{4x^4 + 3x^2 + 4x + 5}{2x^4 + 3x^2 + 4} = 2$

Do you want further details or have additional questions?

Related Questions:

1. How does the degree of a polynomial affect the behavior of a rational function at infinity?
2. Can this method be applied to limits where the degrees of the numerator and denominator differ?
3. What happens if the degree of the denominator is higher than the numerator?
4. How do asymptotes relate to limits at infinity for rational functions?
5. What if the terms in the numerator and denominator had coefficients with opposite signs?

Tip:

For limits of rational functions as $x \to \infty$, always compare the highest degree terms in the numerator and denominator.