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

Answer

To determine 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 1: Evaluate the limit as \( x \) approaches 0

We start by finding:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(e^{\sin x} - 1)}{1 - \cos x} \]

### Step 2: Apply L'Hôpital's Rule

As \( x \) approaches 0, both the numerator and denominator approach 0, creating a \( \frac{0}{0} \) indeterminate form. Thus, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator:

- Derivative of the numerator \( x(e^{\sin x} - 1) \):
- Using the product rule: \( (u \cdot v)' = u'v + uv' \)
- Let \( u = x \) and \( v = e^{\sin x} - 1 \)
- \( u' = 1 \)
- \( v' = e^{\sin x} \cdot \cos x \)

Thus, the derivative of the numerator is:

\[ 1 \cdot (e^{\sin x} - 1) + x \cdot e^{\sin x} \cos x \]

- Derivative of the denominator \( 1 - \cos x \):
- The derivative is \( \sin x \)

Now we rewrite the limit:

\[ \lim_{x \to 0} \frac{(e^{\sin x} - 1) + x e^{\sin x} \cos x}{\sin x} \]

### Step 3: Evaluate the limit again

Now we evaluate this limit as \( x \) approaches 0.

1. As \( x \to 0 \), \( e^{\sin x} \) approaches \( e^0 = 1 \), so \( e^{\sin x} - 1 \) approaches 0.
2. \( \sin x \) approaches 0.
3. The term \( x e^{\sin x} \cos x \) also approaches 0.

Thus, we again have a \( \frac{0}{0} \) form, and we can apply L'Hôpital's Rule again.

### Step 4: Apply L'Hôpital's Rule again

Taking derivatives again:

- The new numerator becomes:

\[ \frac{d}{dx} \left( (e^{\sin x} - 1) + x e^{\sin x} \cos x \right) = e^{\sin x} \cos x + e^{\sin x} \cos x + x e^{\sin x} (-\sin x) \]

- The new denominator becomes:

\[ \cos x \]

Now we have:

\[ \lim_{x \to 0} \frac{2e^{\sin x} \cos x + x e^{\sin x} (-\sin x)}{\cos x} \]

### Step 5: Evaluate the limit as \( x \to 0 \)

As \( x \to 0 \):

- \( e^{\sin x} \) approaches 1,
- \( \cos x \) approaches 1,
- \( \sin x \) approaches 0.

Thus, we have:

\[ \lim_{x \to 0} \frac{2 \cdot 1 \cdot 1 + 0}{1} = 2 \]

### Conclusion

Since the limit exists and is equal to 2, we can conclude that for the function to be continuous at \( x = 0 \):

\[ f(0) = 2 \]

### Final Answer

\[ f(0) = 2 \]