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प्रश्न
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
विकल्प
(a) 10/3
(b) 3/10
(c) 6/5
(d) 5/6
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उत्तर
(a) 10/3
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x}\]
\[ = \lim_{x \to 0} \frac{2 \sin^2 x \sin 5x}{x^2 \sin 3x}\]
\[ = \lim_{x \to 0} \frac{\frac{2 \sin^2 x}{x^2} \left( \frac{\sin 5x}{5x} \right) \times 5}{\left( \frac{\sin 3x}{3x} \right)3}\]
\[ = \frac{10 \times 1^2 \times 1}{1 \times 3}\]
\[ = \frac{10}{3}\]
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