Since the function is continuous at \( x = 0 \), \[ \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(a) \] Left-hand limit: \[ \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{1 - \cos 4x}{x^2} \] Using the identity \( 1 - \cos y = 2 \sin^2 \frac{y}{2} \), we rewrite: \[ = \lim\limits_{x \to 0^-} \left( \frac{2 \sin^2 2x}{(2x)^2} \right) \cdot 4 \] Since \( \frac{\sin u}{u} \approx 1 \) as \( u \to 0 \), we get: \[ = \left( \frac{2 \cdot 1^2}{4} \right) \cdot 4 = 8 \] Right-hand limit: \[ \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4} \] Multiplying numerator and denominator by \( \sqrt{(16 + \sqrt{x})} + 4 \): \[ = \lim\limits_{x \to 0^+} \frac{\sqrt{x} (\sqrt{(16 + \sqrt{x})} + 4)}{(16 + \sqrt{x}) - 16} \] \[ = \lim\limits_{x \to 0^+} \frac{\sqrt{x} (\sqrt{(16 + \sqrt{x})} + 4)}{\sqrt{x}} \] Canceling \( \sqrt{x} \): \[ = \lim\limits_{x \to 0^+} (\sqrt{(16 + \sqrt{x})} + 4) \] Since \( \sqrt{x} \to 0 \), we get: \[ = \sqrt{16} + 4 = 4 + 4 = 8 \] Conclusion: Since \( \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = 8 \) and \( f(0) = 8 \), \[ a = 8. \] Thus, the final answer is: \( 8 \)