The user wants me to solve a limit problem with a parameter a and find the value of a and the limit L given that L is finite.

The given limit is: $L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}, \quad a > 0$

Step 1: Analyze the form of the limit

As $x \to 0$, the denominator $x^4 \to 0$. For the numerator, we have $a - \sqrt{a^2 - 0^2} - \frac{0^2}{4} = a - \sqrt{a^2} = a - |a|$. Since $a > 0$, $|a| = a$. Thus, the numerator is $a - a = 0$. The limit is in the indeterminate form $\frac{0}{0}$.

Step 2: Use Taylor Series Expansion

To evaluate the limit, we can use the Taylor series expansion for the term $\sqrt{a^2 - x^2}$ around $x=0$. $\sqrt{a^2 - x^2} = a\sqrt{1 - \frac{x^2}{a^2}} = a \left(1 - \frac{x^2}{a^2}\right)^{1/2}$ Using the binomial expansion $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + O(u^3)$, with $u = -\frac{x^2}{a^2}$ and $n = \frac{1}{2}$: $\left(1 - \frac{x^2}{a^2}\right)^{1/2} = 1 + \frac{1}{2}\left(-\frac{x^2}{a^2}\right) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2}\left(-\frac{x^2}{a^2}\right)^2 + O(x^6)$ $= 1 - \frac{x^2}{2a^2} + \frac{-\frac{1}{4}}{2}\left(\frac{x^4}{a^4}\right) + O(x^6)$ $= 1 - \frac{x^2}{2a^2} - \frac{x^4}{8a^4} + O(x^6)$ Therefore, $\sqrt{a^2 - x^2} = a\left(1 - \frac{x^2}{2a^2} - \frac{x^4}{8a^4} + O(x^6)\right) = a - \frac{x^2}{2a} - \frac{x^4}{8a^3} + O(x^6)$

Step 3: Substitute the expansion into the limit expression

Now, substitute this expansion into the numerator of the limit: Numerator = $a - \left(a - \frac{x^2}{2a} - \frac{x^4}{8a^3} + O(x^6)\right) - \frac{x^2}{4}$ $= a - a + \frac{x^2}{2a} + \frac{x^4}{8a^3} - \frac{x^2}{4} + O(x^6)$ $= \left(\frac{1}{2a} - \frac{1}{4}\right)x^2 + \frac{1}{8a^3}x^4 + O(x^6)$ So the limit becomes: $L = \lim_{x \to 0} \frac{\left(\frac{1}{2a} - \frac{1}{4}\right)x^2 + \frac{1}{8a^3}x^4 + O(x^6)}{x^4}$

Step 4: Find the value of 'a' for a finite limit

The limit expression can be written as: $L = \lim_{x \to 0} \left[ \left(\frac{1}{2a} - \frac{1}{4}\right) \frac{1}{x^2} + \frac{1}{8a^3} + O(x^2) \right]$ For the limit $L$ to be finite, the term with $\frac{1}{x^2}$ must be eliminated. This requires its coefficient to be zero. $\frac{1}{2a} - \frac{1}{4} = 0$ $\frac{1}{2a} = \frac{1}{4} \implies 2a = 4 \implies a = 2$ So, for $L$ to be finite, $a$ must be 2. This confirms option A is correct.

Step 5: Calculate the value of L

With $a=2$, the coefficient of $x^2$ is zero. The limit expression simplifies to: $L = \lim_{x \to 0} \frac{\frac{1}{8a^3}x^4 + O(x^6)}{x^4}$ Substitute $a=2$: $L = \lim_{x \to 0} \frac{\frac{1}{8(2^3)}x^4 + O(x^6)}{x^4} = \lim_{x \to 0} \frac{\frac{1}{8 \times 8}x^4 + O(x^6)}{x^4}$ $L = \lim_{x \to 0} \left( \frac{1}{64} + O(x^2) \right)$ $L = \frac{1}{64}$ This confirms option C is correct.

Step 6: Conclusion

We have found that for the limit $L$ to be finite, $a=2$ and the value of the limit is $L = \frac{1}{64}$.

• Option A: $a=2$ is correct.
• Option B: $a=1$ is incorrect.
• Option C: $L = \frac{1}{64}$ is correct.
• Option D: $L = \frac{1}{32}$ is incorrect.

Thus, the correct options are A and C.