Question:

<p><strong>Step 1:</strong> Recognize this is of the form \(\frac{\infty}{\infty}\), so apply L'Hôpital's rule.</p><p><strong>Step 2:</strong> Differentiate numerator and denominator:</p><p>Numerator: \(\frac{d}{dx}[\log \log(1-x^2)] = \frac{1}{\log(1-x^2)} \cdot \frac{1}{1-x^2} \cdot (-2x) = \frac{-2x}{(1-x^2)\log(1-x^2)}\)</p><p>Denominator: \(\frac{d}{dx}[\log \log \cos x] = \frac{1}{\log \cos x} \cdot \frac{1}{\cos x} \cdot (-\sin x) = \frac{-\sin x}{\cos x \log \cos x}\)</p><p><strong>Step 3:</strong> The limit becomes:</p><p>\[\lim_{x \to 0} \frac{\frac{-2x}{(1-x^2)\log(1-x^2)}}{\frac{-\sin x}{\cos x \log \cos x}} = 2 \lim_{x \to 0} \frac{x \cos x \log \cos x}{\sin x (1-x^2) \log(1-x^2)}\]</p><p><strong>Step 4:</strong> Rewrite and apply standard limits:</p><p>\[= 2 \lim_{x \to 0} \frac{x}{\sin x} \cdot \lim_{x \to 0} \frac{\cos x}{1-x^2} \cdot \lim_{x \to 0} \frac{\log \cos x}{\log(1-x^2)}\]</p><p>\[= 2 \cdot 1 \cdot 1 \cdot \lim_{x \to 0} \frac{\frac{-\sin x}{\cos x}}{\frac{-2x}{1-x^2}} = 2 \lim_{x \to 0} \frac{(1-x^2)\sin x}{2x \cos x} = 2 \cdot \frac{1}{2} = 1\]</p><p>∴ Answer is (b) 1</p>