Let's solve each of the limits step by step based on the given functions $f(x) = x^3$ and $g(x) = \frac{\sqrt{x^2 + 5}}{2x^2}$:

(a) $\lim_{x \to 2} f(x)$

We are given $f(x) = x^3$. So,

$\lim_{x \to 2} f(x) = \lim_{x \to 2} x^3 = 2^3 = 8$

(b) $\lim_{x \to 2} g(x)$

We are given $g(x) = \frac{\sqrt{x^2 + 5}}{2x^2}$. Now, let's compute the limit:

$\lim_{x \to 2} g(x) = \lim_{x \to 2} \frac{\sqrt{x^2 + 5}}{2x^2}$ Substitute $x = 2$:

$g(2) = \frac{\sqrt{2^2 + 5}}{2(2^2)} = \frac{\sqrt{4 + 5}}{2(4)} = \frac{\sqrt{9}}{8} = \frac{3}{8}$

(c) $\lim_{x \to 2} [f(x)g(x)]$

Using the results from parts (a) and (b), we can find this limit as the product of the two previous limits:

$\lim_{x \to 2} [f(x)g(x)] = \lim_{x \to 2} f(x) \cdot \lim_{x \to 2} g(x)$

Substitute the values:

$8 \cdot \frac{3}{8} = 3$

(d) $\lim_{x \to 2} [g(x) - f(x)]$

Now, subtract the limits from parts (a) and (b):

$\lim_{x \to 2} [g(x) - f(x)] = \lim_{x \to 2} g(x) - \lim_{x \to 2} f(x)$

Substitute the values:

$\frac{3}{8} - 8 = \frac{3}{8} - \frac{64}{8} = \frac{-61}{8}$

Final Answers:

(a) $\lim_{x \to 2} f(x) = 8$

(b) $\lim_{x \to 2} g(x) = \frac{3}{8}$

(c) $\lim_{x \to 2} [f(x)g(x)] = 3$

(d) $\lim_{x \to 2} [g(x) - f(x)] = \frac{-61}{8}$

Would you like more details or clarifications on any of these calculations? Here are five related questions you can explore:

1. What happens to the limit of $f(x)$ as $x \to \infty$?
2. How would you compute $\lim_{x \to 0} g(x)$?
3. What techniques could be used if the limits involved indeterminate forms?
4. Can you find the derivative of $g(x)$?
5. What is the behavior of $g(x)$ as $x \to \infty$?

Tip: When working with products of limits, always check if each individual limit exists first!