Question

If \( L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2}}{x^4} \), \( a > 0 \), then:

• A. \( a = 2 \)
• B. \( a = 1 \)
• C. \( a = \frac{1}{3} \)
• D. None of these

Hint:

Use binomial expansion to approximate expressions involving square roots for small values of \( x \).

Answer 1

The Correct Option is A

We are given the limit: \[ L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2}}{x^4} \] Step 1: Apply binomial expansion We first expand \( \sqrt{a^2 - x^2} \) using the binomial expansion for small \( x \): \[ \sqrt{a^2 - x^2} = a \left(1 - \frac{x^2}{a^2}\right)^{1/2} \] For small \( x \), we approximate this expansion as: \[ \sqrt{a^2 - x^2} \approx a \left(1 - \frac{x^2}{2a^2}\right) \] Step 2: Simplify the limit Substitute this approximation into the limit expression: \[ L = \lim_{x \to 0} \frac{a - \left( a - \frac{x^2}{2a} \right)}{x^4} \] Simplify the numerator: \[ L = \lim_{x \to 0} \frac{\frac{x^2}{2a}}{x^4} = \lim_{x \to 0} \frac{1}{2a x^2} \] For \( L \) to be finite as \( x \to 0 \), we must have \( a = 2 \). Thus, the correct answer is \( a = 2 \).