The limit problem depicted in the image is:

$\lim_{{x \to 5}} \frac{\sqrt{x + 4} - 3}{x - 5}$

To solve this limit, we'll first try substituting $x = 5$ directly into the expression.

$\frac{\sqrt{5 + 4} - 3}{5 - 5} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$

This results in an indeterminate form $\frac{0}{0}$, so we need to simplify the expression further. We will rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator:

$\frac{\sqrt{x + 4} - 3}{x - 5} \cdot \frac{\sqrt{x + 4} + 3}{\sqrt{x + 4} + 3} = \frac{(\sqrt{x + 4} - 3)(\sqrt{x + 4} + 3)}{(x - 5)(\sqrt{x + 4} + 3)}$

Using the difference of squares formula for the numerator:

$(\sqrt{x + 4} - 3)(\sqrt{x + 4} + 3) = (x + 4) - 9 = x - 5$

Now, the expression becomes:

$\frac{x - 5}{(x - 5)(\sqrt{x + 4} + 3)}$

We can cancel the $x - 5$ terms:

$\frac{1}{\sqrt{x + 4} + 3}$

Now, we can substitute $x = 5$ into the simplified expression:

$\frac{1}{\sqrt{5 + 4} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Thus, the value of the limit is:

$\lim_{{x \to 5}} \frac{\sqrt{x + 4} - 3}{x - 5} = \frac{1}{6}$

Would you like further details or have any questions?

Here are 5 related questions to further explore the concept:

1. How do you determine if a limit results in an indeterminate form?
2. Why does rationalizing the numerator help solve limits like this one?
3. Can you explain the difference of squares formula used in the simplification process?
4. What are other techniques to solve limits that result in the form $\frac{0}{0}$?
5. What happens if you encounter a limit where the direct substitution does not result in an indeterminate form?

Tip: When you encounter a limit that results in an indeterminate form, always check if algebraic techniques like factoring, rationalizing, or L'Hôpital's Rule can simplify the expression.