To solve the given limit:

$\lim_{x \to 0} \frac{\sqrt{6x + 64} - 8}{5x}$

First, evaluate the expression directly by substituting $x = 0$:

$\frac{\sqrt{6 \cdot 0 + 64} - 8}{5 \cdot 0} = \frac{\sqrt{64} - 8}{0} = \frac{8 - 8}{0} = \frac{0}{0}$

This result is an indeterminate form, so we need to use a different approach such as L'Hopital's Rule, which states that if we have an indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can take the derivative of the numerator and the denominator and then re-evaluate the limit.

The derivative of the numerator $\sqrt{6x + 64} - 8$ using the chain rule is:

$\frac{d}{dx}(\sqrt{6x + 64}) = \frac{1}{2}(6x + 64)^{-1/2} \cdot 6 = \frac{3}{\sqrt{6x + 64}}$

The derivative of the denominator $5x$ is:

$\frac{d}{dx}(5x) = 5$

Now, applying L'Hopital's Rule gives us:

$\lim_{x \to 0} \frac{\frac{3}{\sqrt{6x + 64}}}{5} = \frac{3}{5 \cdot \sqrt{6 \cdot 0 + 64}} = \frac{3}{5 \cdot 8} = \frac{3}{40}$

Thus, the limit is $\frac{3}{40}$.

Would you like further details on this process or do you have any other questions?

Related questions you might consider:

1. What happens if you apply the limit process directly without L'Hopital's rule?
2. How would the limit change if the expression inside the square root was different?
3. Why is the derivative of $\sqrt{6x + 64}$ equal to $\frac{3}{\sqrt{6x + 64}}$?
4. Can you solve this limit using a series expansion instead of L'Hopital's Rule?
5. What other indeterminate forms exist and how can they be solved?

Tip: When using L'Hopital's Rule, always ensure that you are dealing with an indeterminate form before applying the rule, as incorrect application can lead to erroneous results.