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The problem is asking to calculate the following limit:

\[
\lim_{{x \to 0}} \frac{\sqrt{x+1} - 1}{x}
\]

To solve this, we can use the **conjugate method** to simplify the expression.

### Step-by-Step Solution:

1. **Multiply by the conjugate:**
   The expression has a square root in the numerator, so we multiply both the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{x+1} + 1\). This helps eliminate the square root in the numerator.

\[
\lim_{{x \to 0}} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}
\]

2. **Simplify the numerator:**
   Use the difference of squares formula:

\[
(\sqrt{x+1} - 1)(\sqrt{x+1} + 1) = (x+1) - 1 = x
\]

Thus, the expression becomes:

\[
\lim_{{x \to 0}} \frac{x}{x(\sqrt{x+1} + 1)} = \lim_{{x \to 0}} \frac{1}{\sqrt{x+1} + 1}
\]

3. **Substitute \(x = 0\):**
   Now, we can directly substitute \(x = 0\) into the simplified expression:

\[
\frac{1}{\sqrt{0+1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}
\]

### Final Answer:
\[
\lim_{{x \to 0}} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{2}
\]

Would you like further details or clarifications on this solution?

Here are five related questions to expand your understanding:

1. How does multiplying by the conjugate help simplify limits with square roots?
2. What is the difference of squares formula, and how is it used in this problem?
3. Can you apply L'Hopital's rule to solve this limit problem? How would it change the solution?
4. What are other methods for solving indeterminate forms in limits?
5. How would the limit change if the expression inside the square root were more complex?

**Tip:** Always check for indeterminate forms like \( \frac{0}{0} \) before applying algebraic techniques to simplify limits.

Hitunglah lim x→0 (√(x+1) - 1) / x menggunakan teorema-teorema.

Calculate the limit of (√(x+1) - 1) / x as x approaches 0 using the conjugate method. This problem covers key concepts in limits, indeterminate forms, and the difference of squares formula. Suitable for advanced high school math students.

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