Let `f(x)=(1+sinx)^"cosec x"` , the value of f(0) so that f is a continuous function is

Options

A. e
B. `e^(1//2)`
C. `e^2`
D. 1

Answer

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

### Step-by-Step Solution:

1. **Identify the Function**:
\[ f(x) = (1 + \sin x)^{\csc x} \]
We need to find \( f(0) \) such that \( f(x) \) is continuous at \( x = 0 \).

2. **Check 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 start by evaluating the limit:
\[ \lim_{x \to 0} (1 + \sin x)^{\csc x} \]
Here, \( \csc x = \frac{1}{\sin x} \), so we can rewrite the limit as:
\[ \lim_{x \to 0} (1 + \sin x)^{\frac{1}{\sin x}} \]

4. **Recognize the Indeterminate Form**:
As \( x \to 0 \), \( \sin x \to 0 \), which gives us the form \( 1^\infty \). We can use the exponential limit property:
\[ a^b = e^{b \ln a} \]
Thus, we rewrite the limit:
\[ \lim_{x \to 0} (1 + \sin x)^{\frac{1}{\sin x}} = e^{\lim_{x \to 0} \frac{\ln(1 + \sin x)}{\sin x}} \]

5. **Evaluate the Inner Limit**:
Now we need to find:
\[ \lim_{x \to 0} \frac{\ln(1 + \sin x)}{\sin x} \]
Using L'Hôpital's Rule (since both the numerator and denominator approach 0 as \( x \to 0 \)):
- Differentiate the numerator: \( \frac{d}{dx}[\ln(1 + \sin x)] = \frac{\cos x}{1 + \sin x} \)
- Differentiate the denominator: \( \frac{d}{dx}[\sin x] = \cos x \)

Applying L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\ln(1 + \sin x)}{\sin x} = \lim_{x \to 0} \frac{\frac{\cos x}{1 + \sin x}}{\cos x} = \lim_{x \to 0} \frac{1}{1 + \sin x} \]

6. **Evaluate the Limit**:
As \( x \to 0 \), \( \sin x \to 0 \):
\[ \lim_{x \to 0} \frac{1}{1 + \sin x} = \frac{1}{1 + 0} = 1 \]

7. **Final Calculation**:
Thus, we have:
\[ \lim_{x \to 0} (1 + \sin x)^{\frac{1}{\sin x}} = e^{1} = e \]

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

### Final Answer:
The value of \( f(0) \) so that \( f \) is a continuous function is \( e \).