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प्रश्न
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\tan 8x}{\sin 2x} \right]\]
\[\Rightarrow \lim_{x \to 0} \left[ \frac{\tan 8x}{8x} \times \frac{8x}{\frac{\sin 2x}{2x} \times 2x} \right] \left[ \because \lim_{x \to 0} \frac{\tan 8x}{8x} = 1, \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \right]\]
\[ \Rightarrow 4\]
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