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Question

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


Solution

x=1t
Ln=limt0⎢ ⎢ ⎢(1+t4)(1+t42)(1+t4n)tn1n1t⎥ ⎥ ⎥  
y(t)=((1+t4)(1+t42)(1+t4n))1n
ln y(t)=1n{ln(1+t4)+ln(1+t42)++ln(1+t4n}
y(t)y(t)=1n[1t+4+1t+42++(1t+4n)]
Ln=limt01n(1t+4++1t+4n)((1+t4)(1+t4n))1n
Ln=14+142++14nn=1n(14)(1(14)n)(114)
limn1(14)n3n.n=13
So (x1x)(13)=|logex| has two solutions

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