`f(0) = 1/2`, find `lim _(x rarr0) frac{x int_0^x f(t) dt}{e^(x^2) - 1} = alpha` then find `8 alpha^2`.

Options

A. 3
B. 1
C. 2
D. 0

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{x \int_0^x f(t) \, dt}{e^{x^2} - 1} \]

Given that \( f(0) = \frac{1}{2} \), we can proceed step by step.

### Step 1: Analyze the denominator

First, we need to analyze the denominator \( e^{x^2} - 1 \) as \( x \to 0 \). We can use the Taylor series expansion for \( e^{x^2} \):

\[ e^{x^2} = 1 + x^2 + \frac{x^4}{2} + O(x^6) \]

Thus,

\[ e^{x^2} - 1 = x^2 + \frac{x^4}{2} + O(x^6) \]

As \( x \to 0 \), the leading term in the denominator is \( x^2 \).

### Step 2: Analyze the numerator

Next, we analyze the numerator \( x \int_0^x f(t) \, dt \). Since \( f(0) = \frac{1}{2} \), we can approximate \( f(t) \) near \( t = 0 \). Assuming \( f(t) \) is continuous at \( t = 0 \), we can write:

\[ f(t) \approx f(0) = \frac{1}{2} \text{ for small } t \]

Thus, we can approximate the integral:

\[ \int_0^x f(t) \, dt \approx \int_0^x \frac{1}{2} \, dt = \frac{1}{2} x \]

Therefore, the numerator becomes:

\[ x \int_0^x f(t) \, dt \approx x \cdot \frac{1}{2} x = \frac{1}{2} x^2 \]

### Step 3: Substitute into the limit

Now substituting back into the limit, we have:

\[ \lim_{x \to 0} \frac{\frac{1}{2} x^2}{x^2 + O(x^4)} \]

As \( x \to 0 \), the higher-order terms \( O(x^4) \) become negligible compared to \( x^2 \). Thus, we can simplify the limit:

\[ \lim_{x \to 0} \frac{\frac{1}{2} x^2}{x^2} = \frac{1}{2} \]

### Step 4: Conclusion

Thus, we find that:

\[ \alpha = \frac{1}{2} \]

Finally, we need to find \( 8 \alpha^2 \):

\[ 8 \alpha^2 = 8 \left(\frac{1}{2}\right)^2 = 8 \cdot \frac{1}{4} = 2 \]

So, the final answer is:

\[ \boxed{2} \]