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प्रश्न
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{x \cos x + 2 \sin x}{x^2 + \tan x} \right]\]
Dividing the numerator and the denominator by x, we get:
\[\lim_{x \to 0} \left[ \frac{\cos x + \frac{2\sin x}{x}}{x + \frac{\tan x}{x}} \right]\]
\[ = \frac{\cos0 + 2 \times 1}{0 + 1} \left[ \because \lim_{x \to 0} \left( \frac{\sin x}{x} \right) = 1, \lim_{x \to 0} \left( \frac{\tan x}{x} \right) = 1 \right]\]
\[ = \frac{3}{1}\]
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