To evaluate these limits, let's simplify each expression.

a) $\lim_{x \to \infty} (11 + x^{-2})$

As $x \to \infty$, $x^{-2} = \frac{1}{x^2} \to 0$.

So, the limit is: $11 + 0 = 11$

b) $\lim_{x \to \infty} \frac{3x^2 + 11x^5}{2x^7 + 4}$

The highest power of $x$ in the numerator is $x^5$ and in the denominator is $x^7$. Divide every term by $x^7$:

$\frac{\frac{3x^2}{x^7} + \frac{11x^5}{x^7}}{1 + \frac{4}{x^7}} = \frac{\frac{3}{x^5} + \frac{11}{x^2}}{2 + \frac{4}{x^7}}$

As $x \to \infty$, these terms go to 0, yielding: $\frac{0 + 0}{2 + 0} = 0$

c) $\lim_{x \to \infty} \left( \frac{5x^3}{x^3 + 2x^2 + 1} \right)$

Factor out the highest degree term, $x^3$:

$\frac{5x^3}{x^3(1 + \frac{2}{x} + \frac{1}{x^3})} = \frac{5}{1 + \frac{2}{x} + \frac{1}{x^3}}$

As $x \to \infty$, $\frac{2}{x} \to 0$ and $\frac{1}{x^3} \to 0$, so the limit is: $\frac{5}{1 + 0 + 0} = 5$

Therefore, the solutions are:

• a) 11
• b) 0
• c) 5