Since the limit is in the indeterminate form (0)/(0), we can apply L'Hôpital's Rule. This rule states that if lim_x → c (f(x))/(g(x)) is of the form (0)/(0) or (∞)/(∞), then lim_x → c (f(x))/(g(x)) = lim_x → c (f'(x))/(g'(x)) provided the latter limit exists.

Since the limit is in the indeterminate form (0)/(0), we can apply L'Hôpital's Rule. This rule states that if lim_x → c (f(x))/(g(x)) is of the form (0)/(0) or (∞)/(∞), then lim_x → c (f(x))/(g(x)) = lim_x → c (f'(x))/(g'(x)) provided the latter limit exists.

We differentiate the numerator and the denominator. The derivative of e^x² is e^x² · (2x) by the chain rule, and the derivative of - cos(x) is -(- sin(x)) = sin(x). The derivative of x² is 2x.

Now we substitute the derivatives back into the limit expression. We then try to evaluate the limit again by substituting x = 0. We find that the numerator becomes 2(0) e^0² + sin(0) = 0 + 0 = 0, and the denominator becomes 2(0) = 0. This is still an indeterminate form (0)/(0), so we must apply L'Hôpital's Rule again.

Now we substitute the derivatives back into the limit expression. We then try to evaluate the limit again by substituting x = 0. We find that the numerator becomes 2(0) e^0² + sin(0) = 0 + 0 = 0, and the denominator becomes 2(0) = 0. This is still an indeterminate form (0)/(0), so we must apply L'Hôpital's Rule again.

We differentiate the new numerator and denominator. For 2xe^x², we use the product rule: 2e^x² + 2x(e^x² · 2x) = 2e^x² + 4x²e^x². The derivative of sin(x) is cos(x). The derivative of 2x is 2.

Evaluate the limit as x approaches 0 of (e^( $x^2$ ) - cos(x)) / ( $x^2$ ).

Answer: $\frac{3}{2}$

Step-by-step solution

Step 1: Check for indeterminate form

First, we substitute $x = 0$ into the numerator and the denominator to check the form of the limit. We find that both the numerator and the denominator approach 0, resulting in an indeterminate form of $\frac{0}{0}$ . This indicates that L'Hôpital's Rule can be applied.

Step 2: Apply L'Hôpital's Rule

Since the limit is in the indeterminate form $\frac{0}{0}$ , we can apply L'Hôpital's Rule. This rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$ provided the latter limit exists.

Step 3: Differentiate numerator and denominator

We differentiate the numerator and the denominator. The derivative of $e^{x^2}$ is $e^{x^2} \cdot (2x)$ by the chain rule, and the derivative of - $\cos(x)$ is -(- $\sin(x)$ ) = $\sin(x)$ . The derivative of $x^2$ is 2x.

Step 4: Substitute derivatives and re-evaluate limit

Now we substitute the derivatives back into the limit expression. We then try to evaluate the limit again by substituting $x = 0$ . We find that the numerator becomes 2(0) $e^{0^2} + \sin(0) = 0 + 0 = 0$ , and the denominator becomes 2(0) = 0. This is still an indeterminate form $\frac{0}{0}$ , so we must apply L'Hôpital's Rule again.

Step 5: Apply L'Hôpital's Rule again

We differentiate the new numerator and denominator. For $2xe^{x^2}$ , we use the product rule: $2e^{x^2} + 2x(e^{x^2} \cdot 2x) = 2e^{x^2} + 4x^2e^{x^2}$ . The derivative of $\sin(x)$ is $\cos(x)$ . The derivative of 2x is 2.

Step 6: Evaluate the final limit

Finally, we substitute $x = 0$ into the expression with the second derivatives. The numerator becomes $2e^0 + 4(0)^2e^0 + \cos(0) = 2(1) + 0 + 1 = 3$ . The denominator is 2. Thus, the limit is $\frac{3}{2}$ .

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