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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 |
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Answer» 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 |
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