`lim_(x to 0)(int_(0^(x) x e^(t^(2))dt)/(1+x-e^(x))` is equal to

Options

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

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{\int_{0}^{x} x e^{t^2} dt}{1 + x - e^x} \]

we can apply L'Hôpital's rule since both the numerator and denominator approach 0 as \(x\) approaches 0.

### Step 1: Verify the form of the limit
As \(x \to 0\):
- The numerator: \(\int_{0}^{x} x e^{t^2} dt \to 0\)
- The denominator: \(1 + x - e^x \to 1 + 0 - 1 = 0\)

Thus, we have a \(\frac{0}{0}\) form, which allows us to apply L'Hôpital's rule.

### Step 2: Differentiate the numerator and denominator
Using L'Hôpital's rule, we differentiate the numerator and the denominator with respect to \(x\).

**Numerator:**
Using the Leibniz rule for differentiation under the integral sign:

\[ \frac{d}{dx} \left( \int_{0}^{x} x e^{t^2} dt \right) = \int_{0}^{x} e^{t^2} dt + x \cdot e^{x^2} \]

**Denominator:**
The derivative of the denominator \(1 + x - e^x\) is:

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

### Step 3: Rewrite the limit
Now we rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{\int_{0}^{x} e^{t^2} dt + x e^{x^2}}{1 - e^x} \]

### Step 4: Evaluate the limit
As \(x \to 0\):
- The numerator: \(\int_{0}^{0} e^{t^2} dt + 0 \cdot e^{0^2} = 0 + 0 = 0\)
- The denominator: \(1 - e^0 = 1 - 1 = 0\)

We still have a \(\frac{0}{0}\) form, so we apply L'Hôpital's rule again.

### Step 5: Differentiate again
**Numerator:**
Differentiating again:

\[ \frac{d}{dx} \left( \int_{0}^{x} e^{t^2} dt + x e^{x^2} \right) = e^{x^2} + e^{x^2} + 2x^2 e^{x^2} = 2e^{x^2} + 2x^2 e^{x^2} \]

**Denominator:**
Differentiating again:

\[ \frac{d}{dx}(1 - e^x) = -e^x \]

### Step 6: Rewrite the limit again
Now we have:

\[ \lim_{x \to 0} \frac{2e^{x^2} + 2x^2 e^{x^2}}{-e^x} \]

### Step 7: Evaluate the limit
As \(x \to 0\):
- The numerator: \(2e^{0} + 2 \cdot 0^2 \cdot e^{0} = 2 + 0 = 2\)
- The denominator: \(-e^{0} = -1\)

Thus, the limit becomes:

\[ \frac{2}{-1} = -2 \]

### Final Answer
The limit is

\[ \boxed{-2} \]