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Question
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
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Solution
\[\lim_{x \to 0} \left[ \frac{\cos \left( 2x \right) - 1}{\cos x - 1} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{1 - 2 \sin^2 x - 1}{1 - 2 \sin^2 \left( \frac{x}{2} \right) - 1} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin^2 x}{\sin^2 \left( \frac{x}{2} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin^2 x}{x^2} \times \frac{x^2}{\frac{\sin^2 \left( \frac{x}{2} \right)}{\frac{x^2}{4}} \times \frac{x^2}{4}} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{\sin x}{x} \right)^2 \times \frac{1}{\left( \frac{\sin \left( \frac{x}{2} \right)}{\frac{x}{2}} \right)^2} \times 4 \right]\]
\[ = 4\]
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