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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding Continuity**:
For the function to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) \]
This means we need to find \( \lim_{x \to 0} (x+1)^{\cot x} \).

2. **Substituting \( x = 0 \)**:
If we directly substitute \( x = 0 \) into \( f(x) \):
\[ f(0) = (0 + 1)^{\cot(0)} = 1^{\infty} \]
This is an indeterminate form, so we need to evaluate the limit instead.

3. **Setting up the Limit**:
Let:
\[ y = \lim_{x \to 0} (x+1)^{\cot x} \]
Taking the natural logarithm on both sides gives:
\[ \ln y = \lim_{x \to 0} \cot x \cdot \ln(x+1) \]

4. **Using the Cotangent Identity**:
Recall that \( \cot x = \frac{\cos x}{\sin x} \). As \( x \to 0 \), \( \cot x \) approaches infinity, and we can rewrite:
\[ \ln y = \lim_{x \to 0} \frac{\ln(x+1)}{\tan x} \]
This is now in the form \( \frac{0}{0} \).

5. **Applying L'Hôpital's Rule**:
We can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \), we can take the derivative of the numerator and the denominator:
\[ \ln y = \lim_{x \to 0} \frac{\frac{d}{dx}[\ln(x+1)]}{\frac{d}{dx}[\tan x]} \]
The derivative of \( \ln(x+1) \) is \( \frac{1}{x+1} \), and the derivative of \( \tan x \) is \( \sec^2 x \).

6. **Calculating the Limit**:
Thus, we have:
\[ \ln y = \lim_{x \to 0} \frac{\frac{1}{x+1}}{\sec^2 x} = \lim_{x \to 0} \frac{\cos^2 x}{x+1} \]
Substituting \( x = 0 \):
\[ \ln y = \frac{\cos^2(0)}{0 + 1} = \frac{1}{1} = 1 \]

7. **Finding \( y \)**:
Therefore, we find:
\[ y = e^{\ln y} = e^1 = e \]

8. **Conclusion**:
Since \( y = \lim_{x \to 0} f(x) \), we have:
\[ f(0) = e \]

### Final Answer:
The value of \( f(0) \) must be defined as \( e \) for the function to be continuous at \( x = 0 \).

---