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.