The limit problem from the image is:

$\lim_{{x \to -8}} \frac{\sqrt{1 - x} - 3}{8 + x}$

Solution:

1. Direct Substitution: Let's substitute $x = -8$ directly into the function:

   $\frac{\sqrt{1 - (-8)} - 3}{8 + (-8)} = \frac{\sqrt{1 + 8} - 3}{0} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$

   This results in an indeterminate form $\frac{0}{0}$. Therefore, we need to simplify the expression.

2. Rationalizing the Numerator: To simplify the expression, we multiply both the numerator and denominator by the conjugate of the numerator, $\sqrt{1 - x} + 3$:

   $\frac{\sqrt{1 - x} - 3}{8 + x} \cdot \frac{\sqrt{1 - x} + 3}{\sqrt{1 - x} + 3} = \frac{(\sqrt{1 - x})^2 - 3^2}{(8 + x)(\sqrt{1 - x} + 3)}$

   Simplifying the numerator using the difference of squares:

   $= \frac{(1 - x) - 9}{(8 + x)(\sqrt{1 - x} + 3)} = \frac{-x - 8}{(8 + x)(\sqrt{1 - x} + 3)}$

   Notice that $-x - 8 = -(x + 8)$, so the expression becomes:

   $\frac{-(x + 8)}{(x + 8)(\sqrt{1 - x} + 3)}$

   Now, cancel the $(x + 8)$ terms:

   $= \frac{-1}{\sqrt{1 - x} + 3}$

3. Substitute $x = -8$: Now substitute $x = -8$ into the simplified expression:

   $= \frac{-1}{\sqrt{1 - (-8)} + 3} = \frac{-1}{\sqrt{9} + 3} = \frac{-1}{3 + 3} = \frac{-1}{6}$

Thus, the limit is:

$\lim_{{x \to -8}} \frac{\sqrt{1 - x} - 3}{8 + x} = \frac{-1}{6}$

Would you like any further details on this solution?

Here are five related questions to expand your understanding:

1. What is an indeterminate form and how does it affect the calculation of limits?
2. Why is rationalizing the numerator a useful technique in solving limits?
3. Can limits always be simplified by multiplying with conjugates? If not, what other techniques can be used?
4. What are some common strategies for resolving limits that lead to $\frac{0}{0}$?
5. How do limits behave when there are square roots in the numerator or denominator?

Tip: When you encounter the indeterminate form $\frac{0}{0}$, try algebraic manipulations like factoring, conjugate multiplication, or L'Hopital's Rule to simplify the limit.