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.