The value of `lim_(x rarr 0) (1-cos2x)/(e^(x^(2))-e^(x)+x)` is

Options

A. 0
B. 2
C. 4
D. 8

Explanation

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

### Step 1: Check the form of the limit
As \( x \to 0 \):
- The numerator \( 1 - \cos(2x) \to 1 - 1 = 0 \).
- The denominator \( e^{x^2} - e^x + x \to e^0 - e^0 + 0 = 1 - 1 + 0 = 0 \).

Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule.

**Hint for Step 1:** Check if the limit is in the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator.

- The derivative of the numerator \( 1 - \cos(2x) \) is \( 2\sin(2x) \).
- The derivative of the denominator \( e^{x^2} - e^x + x \) is \( 2xe^{x^2} - e^x + 1 \).

Now we rewrite the limit:
\[ \lim_{x \to 0} \frac{2\sin(2x)}{2xe^{x^2} - e^x + 1} \]

**Hint for Step 2:** Differentiate the numerator and denominator separately.

### Step 3: Evaluate the new limit
Now we substitute \( x = 0 \) into the new limit:
- The numerator becomes \( 2\sin(2 \cdot 0) = 2 \cdot 0 = 0 \).
- The denominator becomes \( 2 \cdot 0 \cdot e^{0} - e^{0} + 1 = 0 - 1 + 1 = 0 \).

Again, we have the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

**Hint for Step 3:** Substitute \( x = 0 \) into the new limit to check if it's still indeterminate.

### Step 4: Differentiate again
Differentiate the numerator and denominator again:
- The derivative of the numerator \( 2\sin(2x) \) is \( 4\cos(2x) \).
- The derivative of the denominator \( 2xe^{x^2} - e^x + 1 \) is \( 2e^{x^2} + 4x^2e^{x^2} - e^x \).

Now we rewrite the limit:
\[ \lim_{x \to 0} \frac{4\cos(2x)}{2e^{x^2} + 4x^2e^{x^2} - e^x} \]

**Hint for Step 4:** Differentiate both the numerator and denominator again.

### Step 5: Substitute \( x = 0 \) again
Now substitute \( x = 0 \):
- The numerator becomes \( 4\cos(2 \cdot 0) = 4 \cdot 1 = 4 \).
- The denominator becomes \( 2e^{0} + 4 \cdot 0^2 \cdot e^{0} - e^{0} = 2 - 1 = 1 \).

Now we have:
\[ \lim_{x \to 0} \frac{4}{1} = 4 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{4} \]