In order that the function `f(x)=(x+1)^(cotx)` is continuous at x = 0, f(0) must be defined as

Options

A. 0
B. `1/e`
C. e
D. 1

Answer

To determine the value of \( f(0) \) such that the function \( f(x) = (x + 1)^{\cot x} \) is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function:
\[ f(x) = (x + 1)^{\cot x} \]

2. **Set Up the Continuity Condition**:
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ f(0) = \lim_{x \to 0} f(x) \]

3. **Evaluate the Limit**:
We need to find:
\[ \lim_{x \to 0} (x + 1)^{\cot x} \]
As \( x \) approaches 0, \( \cot x \) approaches \( \infty \) since \( \cot x = \frac{\cos x}{\sin x} \) and \( \sin x \) approaches 0.

4. **Identify the Indeterminate Form**:
The expression \( (x + 1)^{\cot x} \) approaches \( 1^{\infty} \) as \( x \to 0 \). This is an indeterminate form.

5. **Use the Exponential Limit**:
We can rewrite the limit using the exponential function:
\[ (x + 1)^{\cot x} = e^{\cot x \cdot \ln(x + 1)} \]
Therefore, we need to evaluate:
\[ \lim_{x \to 0} \cot x \cdot \ln(x + 1) \]

6. **Simplify \( \cot x \)**:
Recall that:
\[ \cot x = \frac{\cos x}{\sin x} \]
As \( x \to 0 \), we can use the approximation \( \sin x \approx x \) and \( \cos x \approx 1 \):
\[ \cot x \approx \frac{1}{x} \]

7. **Evaluate \( \ln(x + 1) \)**:
Using the Taylor expansion for \( \ln(1 + x) \) around \( x = 0 \):
\[ \ln(x + 1) \approx x \quad \text{as } x \to 0 \]

8. **Combine the Results**:
Thus, we have:
\[ \cot x \cdot \ln(x + 1) \approx \frac{1}{x} \cdot x = 1 \]

9. **Final Limit Calculation**:
Therefore:
\[ \lim_{x \to 0} \cot x \cdot \ln(x + 1) = 1 \]
Hence:
\[ \lim_{x \to 0} e^{\cot x \cdot \ln(x + 1)} = e^1 = e \]

10. **Conclusion**:
Therefore, for \( f(x) \) to be continuous at \( x = 0 \), we must define:
\[ f(0) = e \]

### Final Answer:
\[ f(0) \text{ must be defined as } e. \]