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

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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}$.