`lim_(xrarr0)((1-e^x)sin x )/(x^2+x^3)` is equal to

Options

A. `-1`
B. `0`
C. `1`
D. `2`

Explanation

To solve the limit \( \lim_{x \to 0} \frac{(1 - e^x) \sin x}{x^2 + x^3} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{(1 - e^x) \sin x}{x^2 + x^3} \]

### Step 2: Factor the denominator
We can factor the denominator:
\[ x^2 + x^3 = x^2(1 + x) \]
Thus, the limit becomes:
\[ \lim_{x \to 0} \frac{(1 - e^x) \sin x}{x^2(1 + x)} \]

### Step 3: Separate the limit
Now, we can separate the limit into two parts:
\[ \lim_{x \to 0} \frac{(1 - e^x)}{x^2} \cdot \frac{\sin x}{1 + x} \]

### Step 4: Evaluate the limits
1. **Evaluate \( \lim_{x \to 0} \frac{1 - e^x}{x^2} \)**:
Using the Taylor series expansion for \( e^x \) around \( x = 0 \):
\[ e^x \approx 1 + x + \frac{x^2}{2} + O(x^3) \]
Therefore,
\[ 1 - e^x \approx -x - \frac{x^2}{2} + O(x^3) \]
So,
\[ \frac{1 - e^x}{x^2} \approx \frac{-x - \frac{x^2}{2}}{x^2} = -\frac{1}{x} - \frac{1}{2} \]
As \( x \to 0 \), this limit approaches \( -\frac{1}{2} \).

2. **Evaluate \( \lim_{x \to 0} \frac{\sin x}{1 + x} \)**:
We know that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), thus:
\[ \lim_{x \to 0} \frac{\sin x}{1 + x} = \frac{0}{1} = 0 \]

### Step 5: Combine the limits
Now we combine the limits:
\[ \lim_{x \to 0} \frac{(1 - e^x)}{x^2} \cdot \lim_{x \to 0} \frac{\sin x}{1 + x} = \left(-\frac{1}{2}\right) \cdot 0 = 0 \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to 0} \frac{(1 - e^x) \sin x}{x^2 + x^3} = 0 \]

### Conclusion
The answer is \( 0 \).