If the function `f(x) = (x(e^(sinx) -1))/( 1 - cos x ) ` is continuous at x =0 then f(0)=

Answer

To find the value of \( f(0) \) for the function

\[ f(x) = \frac{x(e^{\sin x} - 1)}{1 - \cos x} \]

and ensure 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. **Identify the limit expression**:
Since \( f(x) \) is continuous at \( x = 0 \), we have:
\[ f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(e^{\sin x} - 1)}{1 - \cos x} \]

2. **Use known limits**:
We know that:
\[ \lim_{x \to 0} \frac{e^{\sin x} - 1}{\sin x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Also, we can use the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \).

3. **Rewrite the limit**:
Substitute \( 1 - \cos x \) with \( 2 \sin^2\left(\frac{x}{2}\right) \):
\[ f(0) = \lim_{x \to 0} \frac{x(e^{\sin x} - 1)}{2 \sin^2\left(\frac{x}{2}\right)} \]

4. **Rearranging the limit**:
We can rearrange the limit:
\[ f(0) = \frac{1}{2} \lim_{x \to 0} \frac{e^{\sin x} - 1}{\sin x} \cdot \frac{\sin x}{x} \cdot \frac{x}{\sin^2\left(\frac{x}{2}\right)} \]

5. **Evaluate each limit**:
- The first limit \( \lim_{x \to 0} \frac{e^{\sin x} - 1}{\sin x} = 1 \).
- The second limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
- The third limit can be simplified using the fact that \( \sin\left(\frac{x}{2}\right) \approx \frac{x}{2} \) as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{x}{\sin^2\left(\frac{x}{2}\right)} = \lim_{x \to 0} \frac{x}{\left(\frac{x}{2}\right)^2} = \lim_{x \to 0} \frac{x}{\frac{x^2}{4}} = \lim_{x \to 0} \frac{4}{x} = 4 \]

6. **Combine the limits**:
Now, substituting these values back into our limit expression:
\[ f(0) = \frac{1}{2} \cdot 1 \cdot 1 \cdot 4 = \frac{4}{2} = 2 \]

Thus, the value of \( f(0) \) is

\[ \boxed{2} \]