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We are asked to find the limit:

\[
\lim_{{x \to -1}} \frac{-3(x+1)}{|x-1| - |2x + 4|}
\]

### Step-by-step solution:

1. **Analyzing the components**:
   - The numerator is straightforward: \( -3(x+1) \).
   - The denominator involves absolute values: \( |x-1| \) and \( |2x+4| \). 

2. **Handling absolute values**:
   - Let's first evaluate these absolute value expressions around \( x = -1 \).

   For \( |x-1| \):
   \[
   |x-1| = |(-1)-1| = |-2| = 2
   \]
   
   For \( |2x + 4| \):
   \[
   2x + 4 = 2(-1) + 4 = -2 + 4 = 2, \quad \therefore |2x + 4| = 2
   \]

   So at \( x = -1 \), both absolute value terms equal 2.

3. **Substitute \( x = -1 \) into the expression**:
   \[
   \frac{-3(x+1)}{|x-1| - |2x+4|} = \frac{-3((-1)+1)}{|(-1)-1| - |2(-1) + 4|}
   \]

   - The numerator becomes \( -3(0) = 0 \).
   - The denominator becomes \( |-2| - |2| = 2 - 2 = 0 \).

   This results in the indeterminate form \( \frac{0}{0} \).

4. **Applying L'Hôpital's Rule**:
   Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This requires differentiating both the numerator and the denominator.

   - The numerator \( f(x) = -3(x+1) \) has derivative \( f'(x) = -3 \).
   - The denominator \( g(x) = |x-1| - |2x + 4| \) requires separate cases for differentiation.

     - For \( x \to -1 \), \( |x-1| = 1 - x \) and \( |2x + 4| = -(2x+4) = -2x-4 \) (since \( x+2 \) is negative when \( x < -2 \)).
   
     So the derivative of the denominator is:
     \[
     g'(x) = -1 - 2 = -3
     \]

5. **Compute the limit using the derivatives**:
   Applying L'Hôpital's Rule:
   \[
   \lim_{{x \to -1}} \frac{f'(x)}{g'(x)} = \frac{-3}{-3} = 1
   \]

### Final Answer:
\[
\lim_{{x \to -1}} \frac{-3(x+1)}{|x-1| - |2x + 4|} = 1
\]

Would you like more details or have any questions?

### Related Questions:
1. How does L'Hôpital's Rule handle indeterminate forms like \( \frac{0}{0} \)?
2. What are the conditions under which absolute values change behavior when approaching limits?
3. Can this method be applied to other types of indeterminate forms, such as \( \infty/\infty \)?
4. What happens if the limit approaches a point outside the domain of the absolute value expressions?
5. What would the limit be if \( x \) approached \( 1 \) instead of \( -1 \)?

### Tip:
Always check if your limit results in an indeterminate form, such as \( \frac{0}{0} \) or \( \infty/\infty \), before applying L'Hôpital’s Rule.

Find the limit, if it exists. Otherwise, show that the limit does not exist: lim x -> -1 (-3(x+1)) / (|x-1| - |2x+4|).

Find the limit of (-3(x+1)) / (|x-1| - |2x+4|) as x approaches -1. This problem involves understanding limits, absolute values, and applying L'Hôpital's Rule due to the indeterminate form 0/0. A step-by-step solution is provided, suitable for students at an advanced high school or early university level.

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