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प्रश्न
Evaluate the following limits:
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
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उत्तर
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
\[ = \lim_{x \to 0} \frac{- 2\sin\left( \frac{ax + bx}{2} \right)\sin\left( \frac{ax - bx}{2} \right)}{- 2 \sin^2 \frac{cx}{2}} \left( 1 - \cos2\theta = 2 \sin^2 \theta \right)\]
\[ = \lim_{x \to 0} \frac{\left( \frac{a + b}{2} \right)x \times \frac{\sin\left( \frac{a + b}{2} \right)x}{\left( \frac{a + b}{2} \right)x} \times \left( \frac{a - b}{2} \right)x \times \frac{\sin\left( \frac{a - b}{2} \right)x}{\left( \frac{a - b}{2} \right)x}}{\frac{c^2 x^2}{4} \times \frac{\sin^2 \frac{cx}{2}}{\frac{c^2 x^2}{4}}}\]
\[ = \frac{\left( a + b \right)\left( a - b \right)}{c^2} \times \frac{\lim_{x \to 0} \frac{\sin\left( \frac{a + b}{2} \right)x}{\left( \frac{a + b}{2} \right)x} \times \lim_{x \to 0} \frac{\sin\left( \frac{a - b}{2} \right)x}{\left( \frac{a - b}{2} \right)x}}{\left( \lim_{x \to 0} \frac{\sin\frac{cx}{2}}{\frac{cx}{2}} \right)^2}\]
\[= \frac{a^2 - b^2}{c^2} \times \frac{1 \times 1}{1} \left( \lim_\theta \to 0 \frac{\sin\theta}{\theta} = 1 \right)\]
\[ = \frac{a^2 - b^2}{c^2}\]
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