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प्रश्न
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{3 \sin \left( 2x \right) + 2x}{3x + 2 \tan \left( 3x \right)} \right]\]
\[= \lim_{x \to 0} \left[ \frac{\frac{3 \sin 2x}{x} + 2}{3 + \frac{2 \tan 3x}{x}} \right]\]
\[ = \frac{3 \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \right) \times 2 + 2}{3 + 2 \lim_{x \to 0} \left( \frac{\tan 3x}{x} \right) \times 3}\]
\[ = \frac{6 + 2}{3 + 6}\]
\[ = \frac{8}{9}\]
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