To solve the given mathematical expression, we will evaluate each part step by step.

Expression to Evaluate:

$\left( \lim_{x \to \infty} \frac{15x^2 + 5x}{5x^2 + 4x} \right)^{10} + \left( \lim_{x \to 11} \frac{x^2 - 16}{x + 4} \right)^3 + \sum_{k=0}^{\infty} \left( \frac{1018}{1019} \right)^k$

Step 1: Evaluate the first limit

$\lim_{x \to \infty} \frac{15x^2 + 5x}{5x^2 + 4x}$

1. Divide the numerator and the denominator by $x^2$: $\frac{15x^2 + 5x}{5x^2 + 4x} = \frac{15 + \frac{5}{x}}{5 + \frac{4}{x}}$
2. As $x \to \infty$, $\frac{5}{x} \to 0$ and $\frac{4}{x} \to 0$: $\frac{15 + 0}{5 + 0} = \frac{15}{5} = 3$ Thus: $\lim_{x \to \infty} \frac{15x^2 + 5x}{5x^2 + 4x} = 3$ The expression becomes $3^{10}$.

Step 2: Evaluate the second limit

$\lim_{x \to 11} \frac{x^2 - 16}{x + 4}$

1. Factor the numerator: $x^2 - 16 = (x - 4)(x + 4)$
2. The limit becomes: $\lim_{x \to 11} \frac{(x - 4)(x + 4)}{x + 4} = \lim_{x \to 11} (x - 4) = 11 - 4 = 7$ Therefore: $\left( \lim_{x \to 11} \frac{x^2 - 16}{x + 4} \right)^3 = 7^3$

Step 3: Evaluate the infinite sum

$\sum_{k=0}^{\infty} \left( \frac{1018}{1019} \right)^k$

1. This is a geometric series with the first term $a = 1$ and common ratio $r = \frac{1018}{1019}$: $\sum_{k=0}^{\infty} r^k = \frac{1}{1 - r} \quad \text{for} \quad |r| < 1$
2. Calculate the sum: $\frac{1}{1 - \frac{1018}{1019}} = \frac{1}{\frac{1}{1019}} = 1019$

Step 4: Combine all results

$3^{10} + 7^3 + 1019$

1. Calculate $3^{10}$: $3^{10} = 59049$

2. Calculate $7^3$: $7^3 = 343$

3. Combine: $59049 + 343 + 1019 = 60411$

Thus, the final answer is: $\boxed{60411}$

Would you like more details or have any questions?

Questions You Might Ask:

1. How do you evaluate limits involving infinity?
2. What are the properties of geometric series and how are they used?
3. Can you explain factoring techniques for polynomials?
4. What are common simplification strategies for rational functions?
5. How can we identify and simplify limits involving trigonometric functions?

Tip:

For infinite limits, dividing by the highest power of $x$ in the denominator often simplifies the problem significantly.