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The limit presented is:

\[
\lim_{{x \to 3}} \frac{\sqrt{x + 1} - 2}{x - 3}
\]

To solve this limit, we can notice that direct substitution of \(x = 3\) gives an indeterminate form \( \frac{0}{0} \). So, we need to simplify the expression to resolve this indeterminate form. A common approach is to multiply both the numerator and the denominator by the conjugate of the numerator.

### Step-by-Step Solution:

1. **Multiply by the conjugate:**
   Multiply the numerator and denominator by the conjugate of \( \sqrt{x + 1} - 2 \), which is \( \sqrt{x + 1} + 2 \):
   
   \[
   \frac{\sqrt{x + 1} - 2}{x - 3} \cdot \frac{\sqrt{x + 1} + 2}{\sqrt{x + 1} + 2} = \frac{(\sqrt{x + 1} - 2)(\sqrt{x + 1} + 2)}{(x - 3)(\sqrt{x + 1} + 2)}
   \]

2. **Simplify the numerator:**
   Using the difference of squares formula \( (a - b)(a + b) = a^2 - b^2 \), we get:

   \[
   (\sqrt{x + 1})^2 - 2^2 = (x + 1) - 4 = x - 3
   \]

   So, the expression becomes:

   \[
   \frac{x - 3}{(x - 3)(\sqrt{x + 1} + 2)}
   \]

3. **Cancel the common factor:**
   Now, cancel out the \(x - 3\) terms in both the numerator and denominator:

   \[
   \frac{1}{\sqrt{x + 1} + 2}
   \]

4. **Substitute \(x = 3\):**
   Now that we have simplified the expression, we can substitute \(x = 3\) into the remaining expression:

   \[
   \frac{1}{\sqrt{3 + 1} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
   \]

### Final Answer:

\[
\lim_{{x \to 3}} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{1}{4}
\]

Would you like further details or clarifications?

### Related Questions:
1. How do you handle limits that result in an indeterminate form?
2. What are some other techniques for evaluating limits involving square roots?
3. How does multiplying by the conjugate help in simplifying limits?
4. Can L'Hopital's Rule be used for this limit, and how does it work?
5. What are some common mistakes to avoid when simplifying limits?

### Tip:
Always check for indeterminate forms before simplifying limits, as it guides which method (e.g., factoring, conjugates, L'Hopital's Rule) to use.

Learn how to solve the limit problem involving the square root and an indeterminate form. This solution includes multiplying by the conjugate and simplifying the expression, suitable for high school calculus students.

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