`underset(x to 0)lim (e^(x^(2))-cos x)/(sin^(2) x)` is equal to :

Options

A. 2
B. 3
C. `3/2`
D. `5/4`

Answer

To solve the limit

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

we can follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( x = 0 \) into the expression:

\[ \frac{e^{0^2} - \cos(0)}{\sin^2(0)} = \frac{1 - 1}{0} = \frac{0}{0} \]

This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

### 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 \( e^{x^2} - \cos x \) is:
\[ \frac{d}{dx}(e^{x^2}) - \frac{d}{dx}(\cos x) = 2x e^{x^2} + \sin x \]

- The derivative of the denominator \( \sin^2 x \) is:
\[ \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin(2x) \]

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{2x e^{x^2} + \sin x}{\sin(2x)} \]

### Step 3: Evaluate the limit again
Substituting \( x = 0 \) again gives:

\[ \frac{2(0)e^{0^2} + \sin(0)}{\sin(2 \cdot 0)} = \frac{0 + 0}{0} = \frac{0}{0} \]

We apply L'Hôpital's Rule again.

### Step 4: Differentiate again
We differentiate the numerator and the denominator again:

- The derivative of the numerator \( 2x e^{x^2} + \sin x \) is:
\[ 2e^{x^2} + 4x^2 e^{x^2} + \cos x \]

- The derivative of the denominator \( \sin(2x) \) is:
\[ 2\cos(2x) \]

Now we rewrite the limit:

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

### Step 5: Evaluate the limit one more time
Substituting \( x = 0 \):

\[ \frac{2e^{0^2} + 4(0^2)e^{0^2} + \cos(0)}{2\cos(0)} = \frac{2 \cdot 1 + 0 + 1}{2 \cdot 1} = \frac{3}{2} \]

Thus, the final result is:

\[ \lim_{x \to 0} \frac{e^{x^2} - \cos x}{\sin^2 x} = \frac{3}{2} \]