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प्रश्न
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin 3x}{5x} \right]\]
=\[\frac{1}{5} \lim_{x \to 0} \left[ \frac{\sin3x}{3x} \times 3 \right]\]
\[\left[ \because \lim_{x \to 0} \left( \frac{\sin3x}{3x} \right) = 1 \right]\]
= \[\frac{1}{5} \times 1 \times 3\]
= \[\frac{3}{5}\]
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