To evaluate this limit as $x \to \infty$:

$\lim_{x \to \infty} \frac{(3x^4 + 1)(x^2 + 3)}{5x^5 + 7x^2 + 8}$

we'll analyze the degrees of the polynomial in the numerator and the denominator.

Step-by-Step Solution

1. Expand the numerator: The numerator is $(3x^4 + 1)(x^2 + 3)$. Let's expand it: $(3x^4 + 1)(x^2 + 3) = 3x^6 + 9x^4 + x^2 + 3$

   So, the numerator becomes: $3x^6 + 9x^4 + x^2 + 3$

2. Identify the highest degree term in the numerator: The highest degree term in the expanded numerator is $3x^6$.

3. Identify the highest degree term in the denominator: The denominator is $5x^5 + 7x^2 + 8$, where the highest degree term is $5x^5$.

4. Simplify the limit by dividing by the highest power of $x$ in the denominator: $\lim_{x \to \infty} \frac{3x^6 + 9x^4 + x^2 + 3}{5x^5 + 7x^2 + 8}$

   Divide both the numerator and the denominator by $x^5$: $= \lim_{x \to \infty} \frac{3x + \frac{9}{x} + \frac{1}{x^3} + \frac{3}{x^5}}{5 + \frac{7}{x^3} + \frac{8}{x^5}}$

5. Evaluate the limit: As $x \to \infty$, all terms with $\frac{1}{x^k}$ (where $k > 0$) will approach zero. This leaves: $= \frac{3x}{5} \to \infty$

Conclusion

The limit diverges to infinity: $\lim_{x \to \infty} \frac{(3x^4 + 1)(x^2 + 3)}{5x^5 + 7x^2 + 8} = \infty$

Would you like a deeper explanation or have any questions?

Here are five related questions to consider:

1. How do we determine the behavior of polynomials as $x \to \infty$?
2. Why do terms with lower powers of $x$ become negligible in limits as $x \to \infty$?
3. What are some strategies to find limits at infinity for rational functions?
4. How would the result change if the highest degree of $x$ in the denominator was higher than in the numerator?
5. How does L'Hôpital's Rule apply in limits at infinity when both numerator and denominator approach infinity?

Tip: For rational functions, focusing on the highest power of $x$ in the numerator and denominator can simplify limits as $x \to \infty$.