Wednesday, November 9, 2016

limits - Proving limxtoinftyfraclnxxr=0 and limxto0+xrlnx=0 for r>0




This is the question I'm trying to answer



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This is how I went about proving i) and ii):



i) limxlnxxr=limx(1xr1lnxx)=limx(1x11xrlnxx)=limxxlimx1xrlimxlnxx=10=10=0



ii) limx0+xrlnx=limx0+(xr1xlnx)=limx0+xr1limx0+xlnx=limx1xr1limx0+xlnx=limx1x1limx1xrlimx0+xlnx=limxxlimx1xrlimx0+xlnx=10=10=0




For i) I wasn't sure if writing limxx as and limx1xr as 1 so that they cancelled out to give 1 was the right thing to do since the limits for x and xr are different as x gets very large, although they both tend to infinity.
For ii) I wasn't sure if writing limx0+xr1 as limx1xr1 was right to do even though they both give the value 0.


Answer



Your question has already given a big hint. You may assume that these results hold for r=1. Thus we are given the following limits limxlogxx=0=limx0+xlogx With these as given the problem is too simple. We have limxlogxxr=1rlimxlogxrxr=1rlimtlogtt=1r0=0 We have used the substitution t=xr and it is given that t as x. Similarly the second limit is evaluated.


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