Evaluate `lim_(xto0) (e^(x^(2))-cosx)/(x^(2))`

Explanation

To evaluate the limit

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

we first notice that substituting \(x = 0\) directly into the expression results in the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the derivative of the denominator separately.

### Step 1: Apply L'Hôpital's Rule

We differentiate the numerator and denominator:

- **Numerator**: The derivative of \(e^{x^2}\) is \(e^{x^2} \cdot 2x\) (using the chain rule), and the derivative of \(-\cos x\) is \(\sin x\). Therefore, the derivative of the numerator is:

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

- **Denominator**: The derivative of \(x^2\) is \(2x\).

Now we can rewrite our limit:

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

### Step 2: Substitute \(x = 0\)

Substituting \(x = 0\) into the new limit gives us:

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

This is still an indeterminate form, so we apply L'Hôpital's Rule again.

### Step 3: Differentiate Again

We differentiate the numerator and denominator again:

- **Numerator**: For \(e^{x^2} \cdot 2x + \sin x\):

Using the product rule on \(e^{x^2} \cdot 2x\):

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

The derivative of \(\sin x\) is \(\cos x\). Therefore, the new numerator is:

\[ 2e^{x^2}(2x + 1) + \cos x \]

- **Denominator**: The derivative of \(2x\) is \(2\).

Now we rewrite our limit again:

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

### Step 4: Substitute \(x = 0\) Again

Substituting \(x = 0\) gives us:

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

### Final Result

Thus, the limit evaluates to:

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