Evaluate the following limits :
`lim_(x to 0)(log(1+x))/(e^(x)-1)`.

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\log(1+x)}{e^x - 1}, \]

we first check the form of the limit as \(x\) approaches 0.

1. **Substituting \(x = 0\)**:
- The numerator becomes \(\log(1+0) = \log(1) = 0\).
- The denominator becomes \(e^0 - 1 = 1 - 1 = 0\).
- Thus, we have a \( \frac{0}{0} \) indeterminate form.

2. **Applying L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

- Differentiate the numerator:
\[ \frac{d}{dx}[\log(1+x)] = \frac{1}{1+x}. \]

- Differentiate the denominator:
\[ \frac{d}{dx}[e^x - 1] = e^x. \]

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

4. **Substituting \(x = 0\) again**:
- The new numerator becomes \(\frac{1}{1+0} = 1\).
- The new denominator becomes \(e^0 = 1\).
- Thus, we have:
\[ \lim_{x \to 0} \frac{1}{1} = 1. \]

5. **Final Answer**:
Therefore, the limit is:
\[ \boxed{1}. \]