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 find the limit

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{e^{x^2} - e^x + x}, \]

we will use L'Hôpital's Rule since both the numerator and denominator approach 0 as \( x \) approaches 0, resulting in the indeterminate form \( \frac{0}{0} \).

### Step 1: Differentiate the numerator and denominator

The numerator is \( 1 - \cos(2x) \). The derivative of this with respect to \( x \) is:

\[ \frac{d}{dx}(1 - \cos(2x)) = 2\sin(2x). \]

The denominator is \( e^{x^2} - e^x + x \). The derivative of this with respect to \( x \) is:

\[ \frac{d}{dx}(e^{x^2} - e^x + x) = 2xe^{x^2} - e^x + 1. \]

### Step 2: Apply L'Hôpital's Rule

Now we apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{e^{x^2} - e^x + x} = \lim_{x \to 0} \frac{2\sin(2x)}{2xe^{x^2} - e^x + 1}. \]

### Step 3: Evaluate the limit again

As \( x \to 0 \), the numerator \( 2\sin(2x) \) approaches \( 0 \) and the denominator \( 2xe^{x^2} - e^x + 1 \) also approaches \( 0 - 1 + 1 = 0 \). We still have an indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again

We differentiate the numerator and denominator again:

- The derivative of \( 2\sin(2x) \) is \( 4\cos(2x) \).
- The derivative of \( 2xe^{x^2} - e^x + 1 \) is:

\[ \frac{d}{dx}(2xe^{x^2}) = 2e^{x^2} + 4x^2e^{x^2} - e^x. \]

### Step 5: Apply L'Hôpital's Rule again

Now we have:

\[ \lim_{x \to 0} \frac{4\cos(2x)}{2e^{x^2} + 4x^2e^{x^2} - e^x}. \]

### Step 6: Evaluate the limit

Substituting \( x = 0 \):

- The numerator becomes \( 4\cos(0) = 4 \).
- The denominator becomes \( 2e^0 + 0 - e^0 = 2 - 1 = 1 \).

Thus, we have:

\[ \lim_{x \to 0} \frac{4}{1} = 4. \]

### Final Answer

So, the value of the limit is

\[ \boxed{4}. \]