The value of `lim_(x to 0) (e^x-e^sinx)/(2(x-sinx))`, is

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. `1`
D. `(3)/(2)`

Explanation

To solve the limit \( \lim_{x \to 0} \frac{e^x - e^{\sin x}}{2(x - \sin x)} \), we can use Taylor series expansions for \( e^x \) and \( e^{\sin x} \) around \( x = 0 \).

### Step-by-Step Solution:

1. **Expand \( e^x \)**:
The Taylor series expansion of \( e^x \) around \( x = 0 \) is:
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) \]

2. **Expand \( e^{\sin x} \)**:
First, we need the expansion of \( \sin x \):
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
Now substituting \( \sin x \) into the expansion of \( e^x \):
\[ e^{\sin x} = e^{x - \frac{x^3}{6} + O(x^5)} = 1 + \left(x - \frac{x^3}{6}\right) + \frac{(x - \frac{x^3}{6})^2}{2} + O((x - \frac{x^3}{6})^3) \]
For simplicity, we can neglect higher order terms and approximate:
\[ e^{\sin x} \approx 1 + \left(x - \frac{x^3}{6}\right) + \frac{x^2}{2} + O(x^4) \]

3. **Substituting the expansions**:
Now substituting these expansions into the limit:
\[ e^x - e^{\sin x} \approx \left(1 + x + \frac{x^2}{2}\right) - \left(1 + \left(x - \frac{x^3}{6}\right) + \frac{x^2}{2}\right) \]
Simplifying this gives:
\[ e^x - e^{\sin x} \approx x + \frac{x^3}{6} - x = \frac{x^3}{6} \]

4. **Expand \( x - \sin x \)**:
We also need to expand \( x - \sin x \):
\[ x - \sin x = x - \left(x - \frac{x^3}{6} + O(x^5)\right) = \frac{x^3}{6} + O(x^5) \]

5. **Substituting into the limit**:
Now substituting back into the limit:
\[ \lim_{x \to 0} \frac{e^x - e^{\sin x}}{2(x - \sin x)} = \lim_{x \to 0} \frac{\frac{x^3}{6}}{2 \cdot \frac{x^3}{6}} = \lim_{x \to 0} \frac{\frac{x^3}{6}}{\frac{x^3}{3}} = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} \]

### Final Answer:
The value of the limit is:
\[ \frac{1}{2} \]