The limit you provided is:

$\lim_{{x \to 3}} \frac{|x + 2| - 1}{3 + x}$

Solution:

1. Substitute $x = 3$ directly into the expression: $\frac{|3 + 2| - 1}{3 + 3} = \frac{|5| - 1}{6} = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}$

Since the absolute value function does not affect the expression for $x = 3$, this is a straightforward substitution.

Final Answer:

$\lim_{{x \to 3}} \frac{|x + 2| - 1}{3 + x} = \frac{2}{3}$

Would you like more details or have any questions about the steps? Here are 5 follow-up questions for deeper understanding:

1. How does the absolute value function affect limits when the expression inside it changes sign?
2. What happens if we approach a point where the expression inside the absolute value equals zero?
3. How does direct substitution help in limit calculations, and when does it fail?
4. Could this limit be computed differently using piecewise analysis if the absolute value was more complex?
5. What if the limit was evaluated at a point where the denominator equals zero—how would we handle that?

Tip: Always check if direct substitution works for a limit before trying more advanced techniques!