Question:

<p><strong>Step 1: Find limit a</strong></p><p>$$a = \lim_{x \to 0} \frac{\ln(\cos 2x)}{3x^2}$$</p><p>Using Taylor expansion: $\cos 2x = 1 - \frac{(2x)^2}{2} + O(x^4) = 1 - 2x^2 + O(x^4)$</p><p>Then: $\ln(\cos 2x) = \ln(1 - 2x^2 + O(x^4)) = -2x^2 - \frac{(2x^2)^2}{2} + O(x^6) = -2x^2 - 2x^4 + O(x^6)$</p><p>Therefore: $$a = \lim_{x \to 0} \frac{-2x^2 + O(x^4)}{3x^2} = \frac{-2}{3}$$</p><p><strong>Step 2: Find limit b</strong></p><p>$$b = \lim_{x \to 0} \frac{\sin^2 2x}{x(1-e^x)}$$</p><p>Using Taylor expansions: $\sin 2x = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x + O(x^3)$</p><p>So: $\sin^2 2x = 4x^2 + O(x^4)$</p><p>And: $e^x = 1 + x + \frac{x^2}{2} + O(x^3)$, so $1 - e^x = -x - \frac{x^2}{2} + O(x^3)$</p><p>Therefore: $$b = \lim_{x \to 0} \frac{4x^2 + O(x^4)}{x(-x - \frac{x^2}{2} + O(x^3))} = \lim_{x \to 0} \frac{4x^2}{-x^2 + O(x^3)} = \frac{4}{-1} = -4$$</p><p><strong>Step 3: Find limit c</strong></p><p>$$c = \lim_{x \to 1} \frac{x - \sqrt{x}}{\ln x}$$</p><p>Let $u = x - 1$, so as $x \to 1$, $u \to 0$. Then $x = 1 + u$ and $\sqrt{x} = \sqrt{1+u} = 1 + \frac{u}{2} - \frac{u^2}{8} + O(u^3)$</p><p>Numerator: $x - \sqrt{x} = (1+u) - (1 + \frac{u}{2} + O(u^2)) = \frac{u}{2} + O(u^2)$</p><p>Denominator: $\ln x = \ln(1+u) = u - \frac{u^2}{2} + O(u^3)$</p><p>Therefore: $$c = \lim_{u \to 0} \frac{\frac{u}{2} + O(u^2)}{u + O(u^2)} = \frac{1/2}{1} = \frac{1}{2}$$</p><p><strong>Step 4: Compare values</strong></p><p>We have: $a = -\frac{2}{3} \approx -0.667$, $b = -4$, $c = \frac{1}{2} = 0.5$</p><p>Ordering: $b = -4 < a = -\frac{2}{3} < c = \frac{1}{2}$</p><p>This gives us $b < a < c$</p><p>∴ Answer: D</p>