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प्रश्न
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\cos ax - \cos bx}{\cos cx - \cos dx} \right]\] It is of the form \[\left( \frac{0}{0} \right)\]
\[\Rightarrow \lim_{x \to 0} \left[ \frac{- 2\sin\left( \frac{ax + bx}{2} \right)\sin\left( \frac{ax - bx}{2} \right)}{- 2\sin\left( \frac{cx + dx}{a2} \right)\sin\left( \frac{cx - dx}{2} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\sin\left( \frac{ax + bx}{2} \right)}{\left( \frac{ax + b}{x} \right)} \times \left( \frac{ax + bx}{2} \right) \times \frac{\sin\left( \frac{ax - bx}{2} \right)}{\left( \frac{ax - bx}{2} \right)} \times \left( \frac{ax - bx}{2} \right)}{\frac{\sin\left( \frac{cx + dx}{2} \right)}{\left( \frac{cx + dx}{2} \right)} \times \left( \frac{cx + dx}{2} \right) \times \frac{\sin\left( \frac{cx - dx}{2} \right)}{\left( \frac{cx - dx}{2} \right)} \times \left( \frac{cx - dx}{2} \right)} \right]\]
\[ = 1 \times \lim_{x \to 0} \left[ \frac{\left( \frac{ax + bx}{2} \right)\left( \frac{ax - bx}{2} \right)}{\left( \frac{cx + dx}{2} \right)\left( \frac{cx - dx}{2} \right)} \right]\]
\[ = \lim_{x \to 0} \frac{x^2 \left( a + b \right)\left( a - b \right)}{x^2 \left( c + d \right)\left( c - d \right)}\]
\[ = \frac{a^2 - b^2}{c^2 - d^2}\]
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