Monday, 6 July 2015

calculus - Limit (without series expansion and l'Hôpital's rule)



limxlnx+x2+1x+x21(lnx+1x1)2=18




Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.



Note:
WolframAlpha confirms that the result is 18.


Answer



It is easy to observe that as x the argument of both the logarithms tend to 1 and this is exactly what we need when we are dealing with limit of logarithmic terms. Just add 1 and subtract 1 in the argument of log and proceed as follows:
L=limxlogx+x2+1x+x21(logx+1x1)2=limt0+log1+1+t21+1t2(log1+t1t)2 (putting x=1/t)=limt0+log(1+1+1+t21+1t21){log(1+1+t1t1)}2=limt0+log(1+1+t21t21+1t2){log(1+2t1t)}2=limt0+log(1+1+t21t21+1t2)1+t21t21+1t21+t21t21+1t2{log(1+2t1t)}2=limt0+1+t21t21+1t2{log(1+2t1t)2t1t2t1t}2=limt0+1+t21t21+1t2(1t)24t2=18limt0+1+t21t2t2=18limt0+2t2t2{1+t2+1t2}=18


The technique used above is based on the standard limits and is well explained in the comments to this answer.


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