The given limit expression is:

\[
\lim_{{x \to -1}} \frac{x + 4}{x + 1}
\]

### Step-by-Step Solution:

1. **Substitute \(x = -1\) directly into the expression:**

   Substituting \(x = -1\) into the function:
   \[
   \frac{x + 4}{x + 1} = \frac{-1 + 4}{-1 + 1} = \frac{3}{0}
   \]

2. **Identify the form:**
   The result is \(\frac{3}{0}\), which suggests a division by zero. Since the numerator is not zero, we need to examine the behavior of the function around \(x = -1\).

3. **Check the limit from both sides (Left-hand and Right-hand limits):**
   
   - As \(x \to -1^-\) (from the left), \(x + 1\) is slightly negative, meaning the denominator approaches \(0^-\).
     \[
     \lim_{{x \to -1^-}} \frac{x + 4}{x + 1} = \frac{3}{0^-} = -\infty
     \]
   - As \(x \to -1^+\) (from the right), \(x + 1\) is slightly positive, meaning the denominator approaches \(0^+\).
     \[
     \lim_{{x \to -1^+}} \frac{x + 4}{x + 1} = \frac{3}{0^+} = +\infty
     \]

4. **Conclusion:**
   Since the left-hand and right-hand limits approach negative and positive infinity respectively, the overall limit does **not exist**.

\[
\boxed{\text{The limit does not exist.}}
\]

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. What is the limit of \(\frac{x^2 - 1}{x - 1}\) as \(x \to 1\)?
2. How can we compute limits approaching infinity?
3. What are the rules for evaluating limits at infinity?
4. How do we handle indeterminate forms like \(\frac{0}{0}\)?
5. How does L'Hopital's rule help in calculating limits?

**Tip:** When a function gives division by zero, check whether it leads to a finite limit, infinity, or if the limit does not exist by considering behavior from both sides.

Understand how to solve the limit of (x + 4) / (x + 1) as x approaches -1. This problem explores limits, left-hand and right-hand limits, and the behavior of functions near division by zero. Suitable for Grades 11-12 calculus students.

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