Let L = limx→0 (a - √(a² - x²) - x²/4x⁴), a > 0. If L is finite, then what are the possible values of a and L?

Let the given limit be
L = limx→0 (a - √(a² - x²) - x²/4x⁴)
We can rewrite the expression as:
L = limx→0 [a - a(1 - x²/a²)1/2 - x²/4x⁴]
Using the binomial approximation (1 + x)n ≈ 1 + nx for small x, we have:
(1 - x²/a²)1/2 ≈ 1 - x²/2a²
Substituting this into the expression for L:
L = limx→0 [a - a(1 - x²/2a²) - x²/4x⁴] = limx→0 [a - a + x²/2a - x²/4x⁴] = limx→0 [x²/2a - 1/(4x²)]
For the limit to be finite, the terms with x² in the denominator must cancel out. Let's consider the Taylor expansion of √(a²-x²) around x=0:
√(a²-x²) = a - x²/2a - x⁴/8a³ + O(x⁶)
Substituting this into the original limit expression:
L = limx→0 [a - (a - x²/2a - x⁴/8a³ + O(x⁶)) - x²/4x⁴] = limx→0 [x²/2a + x⁴/8a³ + O(x⁶) - 1/(4x²)]
For the limit to exist, the terms with x⁻² must cancel. This is not possible with the given expression. Let's use L'Hopital's rule. We can rewrite the expression as:
L = limx→0 (a - √(a² - x²) - x²/4x⁴) = limx→0 (a - √(a² - x²) - 1/(4x²))
This limit is of the indeterminate form ∞ - ∞. Let's analyze the terms separately:
limx→0 (a - √(a² - x²)) = 0
limx→0 (-1/(4x²)) = -∞
Therefore, the limit L will be -∞ unless the numerator goes to 0 faster than the denominator. Let's use a more precise approximation:
√(a² - x²) ≈ a(1 - x²/(2a²)) = a - x²/(2a)
Then
L = limx→0 (a - (a - x²/(2a)) - 1/(4x²)) = limx→0 (x²/(2a) - 1/(4x²))
This limit is still -∞. There must be a mistake in the problem statement or a missing term.
If the question intended L = limx→0 (a - √(a² - x²) - x²/4x²) instead, then:
L = limx→0 (x²/(2a) - x²/4) = 0
This implies that the question may contain a typographical error. Without correcting the error, a definitive solution for a and L cannot be provided.