Evaluate `lim_(xto0) (x(e^(x)-1))/(1-cosx)` is equal to

Options

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

Explanation

To evaluate the limit

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

we first substitute \(x = 0\) directly into the expression.

1. **Direct Substitution**:
\[ \frac{0(e^0 - 1)}{1 - \cos(0)} = \frac{0(1 - 1)}{1 - 1} = \frac{0}{0}. \]
This results in the indeterminate form \( \frac{0}{0} \).

**Hint**: When you encounter the \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

- **Numerator**:
\[ \frac{d}{dx}[x(e^x - 1)] = e^x - 1 + xe^x. \]
- **Denominator**:
\[ \frac{d}{dx}[1 - \cos x] = \sin x. \]

Now we rewrite the limit:
\[ \lim_{x \to 0} \frac{e^x - 1 + xe^x}{\sin x}. \]

**Hint**: After differentiating, check if substituting \(x = 0\) again gives an indeterminate form.

3. **Substituting Again**:
Substitute \(x = 0\) into the new limit:
\[ \frac{e^0 - 1 + 0 \cdot e^0}{\sin(0)} = \frac{1 - 1 + 0}{0} = \frac{0}{0}. \]
We still have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

4. **Applying L'Hôpital's Rule Again**:
Differentiate the numerator and denominator again:
- **Numerator**:
\[ \frac{d}{dx}[e^x - 1 + xe^x] = e^x + e^x + xe^x = 2e^x + xe^x. \]
- **Denominator**:
\[ \frac{d}{dx}[\sin x] = \cos x. \]

Now we have:
\[ \lim_{x \to 0} \frac{2e^x + xe^x}{\cos x}. \]

**Hint**: Substitute \(x = 0\) again to see if we can find the limit.

5. **Final Substitution**:
Substitute \(x = 0\):
\[ \frac{2e^0 + 0 \cdot e^0}{\cos(0)} = \frac{2 \cdot 1 + 0}{1} = \frac{2}{1} = 2. \]

Thus, the limit is

\[ \boxed{2}. \]