Question

Define \( f: \mathbb{R} \to \mathbb{R} \) by \[ f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases} \] Find the value of \( a \) such that \( f \) is continuous at \( x = 0 \).

• A. \( 8 \)
• B. \( 4 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

Use the small-angle approximation \( \cos x \approx 1 - \frac{x^2}{2} \) for trigonometric limits. - Multiply by conjugates to simplify square root expressions in limits.

Answer 1

The Correct Option is A

Step 1: Check left-hand limit (\( x \to 0^- \))
Using the standard limit: \[ \lim_{x \to 0} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{x^2}. \] Using \( \sin x \approx x \) for small \( x \): \[ \lim_{x \to 0} \frac{2 (4x)^2}{x^2} = \lim_{x \to 0} \frac{32x^2}{x^2} = 32. \] Thus, \[ \lim_{x \to 0^-} f(x) = 8. \] Step 2: Check right-hand limit (\( x \to 0^+ \))
Rewriting the function: \[ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}. \] Multiply numerator and denominator by the conjugate: \[ \frac{\sqrt{x} (\sqrt{16 + \sqrt{x}} + 4)}{(16 + \sqrt{x}) - 16}. \] \[ = \frac{\sqrt{x} (\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}}. \] \[ = \sqrt{16 + \sqrt{x}} + 4. \] Taking the limit as \( x \to 0 \): \[ \lim_{x \to 0^+} f(x) = \sqrt{16} + 4 = 8. \] Step 3: Compute \( a \)
For continuity: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0). \] Thus, \( a = 8 \), giving the final answer as \( \boxed{8} \).