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Question
\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\]
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Solution
\[\lim_{x \to 0} \left[ \frac{2x - \sin x}{\tan x + x} \right]\] Dividing the numerator and the denominator by x, we get:
\[\lim_{x \to 0} \left[ \frac{2 - \frac{\sin x}{x}}{\frac{\tan x}{x} + 1} \right]\]
\[ = \frac{2 - 1}{1 + 1} \left[ \because \lim_{x \to 0} \frac{\tan x}{x} = 1, \lim_{x \to 0} \frac{\sin x}{x} = 1 \right]\]
\[ = \frac{1}{2}\]
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