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

Options

A. f(0)=0
B. f(0)=e
C. f(0)=1/e
D. none of these

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 find the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

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

2. **Substitute the Function**:
Substitute \( f(x) \) into the limit:
\[ f(0) = \lim_{x \to 0} (x + 1)^{\cot x} \]

3. **Identify the Indeterminate Form**:
As \( x \to 0 \), \( (x + 1) \to 1 \) and \( \cot x \to \infty \). Thus, we have the form \( 1^{\infty} \), which is indeterminate.

4. **Use the Exponential Limit Transformation**:
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:
\[ f(0) = \lim_{x \to 0} e^{\cot x \cdot \ln(x + 1)} \]

5. **Evaluate the Limit Inside the Exponential**:
We need to find:
\[ \lim_{x \to 0} \cot x \cdot \ln(x + 1) \]
We know:
\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \ln(x + 1) \approx x \text{ as } x \to 0 \]

6. **Use Taylor Series Expansion**:
The Taylor series expansion for \( \ln(x + 1) \) around \( x = 0 \) is:
\[ \ln(x + 1) \approx x \quad \text{for small } x \]
Thus:
\[ \cot x \cdot \ln(x + 1) \approx \cot x \cdot x \]

7. **Rewrite \( \cot x \cdot x \)**:
We can express \( \cot x \) as:
\[ \cot x = \frac{x}{\tan x} \quad \Rightarrow \quad \cot x \cdot x = \frac{x^2}{\tan x} \]

8. **Evaluate the Limit**:
As \( x \to 0 \):
\[ \lim_{x \to 0} \frac{x^2}{\tan x} = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0 \]

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

### Conclusion:
Thus, for the function \( f(x) = (x + 1)^{\cot x} \) to be continuous at \( x = 0 \), we must define:
\[ f(0) = 1 \]