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lim(x →π/6) [(3sinx - √3cosx)/(6x - π)] = (A) √3 (B) 1/√3

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lim(x →π/6) [(3sinx - √3cosx)/(6x - π)] =

(A) √3 

(B) 1/√3

(C) - √3 

(D) -1/√3

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Answer is (B) 1/√3

Using L’Hospital’s rule,

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