The expression in the image is:

$\lim_{x \to -8} \frac{\frac{x}{x+6} - 4}{x+8}$

To evaluate this limit algebraically, let's follow these steps:

Step 1: Simplify the Numerator

The numerator is $\frac{x}{x+6} - 4$. Let's combine this into a single fraction:

$\frac{x}{x+6} - 4 = \frac{x - 4(x+6)}{x+6} = \frac{x - 4x - 24}{x+6} = \frac{-3x - 24}{x+6}$

Simplifying further:

$\frac{-3x - 24}{x+6} = \frac{-3(x + 8)}{x+6}$

Step 2: Substitute the Simplified Numerator into the Limit

Now substitute this expression back into the limit:

$\lim_{x \to -8} \frac{\frac{-3(x+8)}{x+6}}{x+8}$

Simplify by canceling $(x+8)$:

$\frac{-3(x+8)}{(x+6)(x+8)} \cdot \frac{1}{x+8} = \frac{-3}{x+6}$

Step 3: Evaluate the Limit

Now, substitute $x = -8$:

$\lim_{x \to -8} \frac{-3}{x+6} = \frac{-3}{-8+6} = \frac{-3}{-2} = \frac{3}{2}$

Thus, the value of the limit is:

$\boxed{\frac{3}{2}}$

Would you like further details or explanations?

Here are some related questions you might find useful:

1. How do you handle limits when the expression has a removable discontinuity?
2. What are common algebraic techniques used to simplify complex fractions in limits?
3. How do you deal with limits approaching infinity or negative infinity?
4. How can L'Hôpital's rule be applied to limits like this one?
5. What happens if the limit does not exist? How can you identify those cases?

Tip: Always look for common factors to cancel in the numerator and denominator when simplifying limits. This often helps to remove discontinuities.