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If Ln=limx→∞((x+14)(x+142)(x+143)……(x+14n))1n−x then the number of solution of equation (x−1x)(limn→∞(nLn))=|logex| is

Answer» If Ln=limx((x+14)(x+142)(x+143)(x+14n))1nx then the number of solution of equation (x1x)(limn(nLn))=|logex| is


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