Finite Limit Evaluation with Parameter and Radical Function

To find the value of $a$ and the limit $L$, we first consider the given limit:
$L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}, \quad a > 0.$

Since $a > 0$, we can factor out $a$ from the square root term:
$\sqrt{a^2 - x^2} = a \sqrt{1 - \frac{x^2}{a^2}} = a \left(1 - \frac{x^2}{a^2}\right)^{1/2}.$

Using the binomial expansion $(1 - u)^{1/2} = 1 - \frac{1}{2}u - \frac{1}{8}u^2 - O(u^3)$ for $|u| < 1$, where $u = \frac{x^2}{a^2}$:
$\sqrt{a^2 - x^2} = a \left(1 - \frac{1}{2}\frac{x^2}{a^2} - \frac{1}{8}\frac{x^4}{a^4} + O(x^6)\right) = a - \frac{x^2}{2a} - \frac{x^4}{8a^3} + O(x^6).$

Substituting this expansion into the numerator:
$\text{Numerator} = a - \left(a - \frac{x^2}{2a} - \frac{x^4}{8a^3} + O(x^6)\right) - \frac{x^2}{4}$
$\text{Numerator} = \left(\frac{1}{2a} - \frac{1}{4}\right)x^2 + \frac{x^4}{8a^3} + O(x^6).$

Now, substituting the numerator back into the limit expression:
$L = \lim_{x \to 0} \left[ \left(\frac{1}{2a} - \frac{1}{4}\right)\frac{1}{x^2} + \frac{1}{8a^3} + O(x^2) \right].$

For $L$ to be finite, the coefficient of the singular term $\frac{1}{x^2}$ must vanish:
$\frac{1}{2a} - \frac{1}{4} = 0 \implies 2a = 4 \implies a = 2.$

Substituting $a = 2$ back into the expression for the limit:
$L = \frac{1}{8a^3} = \frac{1}{8(2)^3} = \frac{1}{8 \times 8} = \frac{1}{64}.$

Therefore, we have:

• $a = 2$ (Option A)
• $L = \frac{1}{64}$ (Option C)

The correct options are A and C.