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प्रश्न
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\tan 3x - 2x}{3x - \sin^2 x} \right]\] Dividing the numerator and the denominator by x:
\[\lim_{x \to 0} \left[ \frac{\frac{\tan 3x}{x} - 2}{3 - \frac{\sin^2 x}{x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\tan 3x}{3x} \times 3 - 2}{3 - \left( \frac{\sin x}{x} \right) \times \sin x} \right]\]
\[ = \frac{3 - 2}{3 - 1 \times 0} \left[ \because \lim_{x \to 0} \frac{\tan x}{x} = 1, \lim_{x \to 0} \frac{\sin x}{x} = 1 \right]\]
\[ = \frac{1}{3}\]
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