`lim_(xrarr0) (sqrt(1+x)-1)/(x)=?`

Answer

To find the limit

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

we start by substituting \(x = 0\):

1. **Substituting \(x = 0\)**:
\[ \frac{\sqrt{1+0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form \( \frac{0}{0} \).

**Hint**: When you encounter a \( \frac{0}{0} \) form, you can use L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

- **Numerator**: \( \sqrt{1+x} - 1 \)
- The derivative of \( \sqrt{1+x} \) is \( \frac{1}{2\sqrt{1+x}} \) (using the chain rule).
- Thus, the derivative of the numerator is:
\[ \frac{d}{dx}(\sqrt{1+x} - 1) = \frac{1}{2\sqrt{1+x}} \cdot 1 = \frac{1}{2\sqrt{1+x}} \]

- **Denominator**: \( x \)
- The derivative of \( x \) is \( 1 \).

3. **Re-evaluating the limit**:
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}}}{1} = \lim_{x \to 0} \frac{1}{2\sqrt{1+x}} \]

4. **Substituting \(x = 0\) again**:
\[ \frac{1}{2\sqrt{1+0}} = \frac{1}{2\sqrt{1}} = \frac{1}{2} \]

Thus, the limit is:

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

### Final Answer:
\[ \frac{1}{2} \]