# How do you evaluate the limit #(2sinx-sin2x)/(xcosx)# as x approaches #0#?

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To evaluate the limit (2sinx-sin2x)/(xcosx) as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (2cosx-2cos2x)/(cosx-xsinx). Substituting x=0 into this expression, we obtain (2-2)/(1-0) = 0. Therefore, the limit of (2sinx-sin2x)/(xcosx) as x approaches 0 is 0.

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