To solve the limit in part (e):

$\lim_{x \to \infty} \left( \frac{5x^2 + 8x - 2}{5x^2 + 3x + 3} \right)^{4x+1},$

Step 1: Simplify the fraction

For large $x$, the highest power of $x$ dominates in both the numerator and denominator. Therefore, divide the numerator and denominator by $x^2$:

$\frac{5x^2 + 8x - 2}{5x^2 + 3x + 3} = \frac{5 + \frac{8}{x} - \frac{2}{x^2}}{5 + \frac{3}{x} + \frac{3}{x^2}}.$

As $x \to \infty$, the terms $\frac{8}{x}, \frac{2}{x^2}, \frac{3}{x}, \frac{3}{x^2}$ approach 0. Thus, the fraction simplifies to:

$\frac{5 + 0 - 0}{5 + 0 + 0} = 1.$

Step 2: Apply the exponent

The limit now becomes:

$\lim_{x \to \infty} \left( 1 \right)^{4x+1}.$

Step 3: Evaluate the form

The expression $1^{4x+1}$ equals 1 for all values of $x$, so the limit is:

$\boxed{1}.$

---

Would you like me to explain any of the steps in more detail? 😊
Here are 5 related questions you might find helpful:

1. How do you simplify limits of rational functions for large $x$?
2. What is the role of dominant terms in polynomial limits?
3. How do you handle exponential expressions in limits?
4. Can you explain L'Hôpital's Rule and when it applies?
5. How do we evaluate limits involving indeterminate forms like $1^\infty$?

Tip: When dealing with rational functions, always simplify by dividing through by the highest power of $x$ in the denominator!