Tuesday, 27 November 2018

real analysis - About the limxrightarrowinfty(lnxx)




Find the limx(lnxx).





We know that lnx=o(x) as x therefore we can guess that the limit will be .



Intuitively x goes to infinity way faster than lnx.



Here it is my formal proof of this:



We have that limxx2lnx=+ thus form the definition, a>0 such that x>2lnx forall x>a.



Now from this,x>a, we deduce that xlnx>2lnxlnx=lnxlnxx<lnx




Finally we have lim supx(lnxx)



Thus lim supx(lnxx)=lim infx(lnxx)= .



Is my argument correct?



Thank you in advance!


Answer



for x>0,




ln(x)x=x(ln(x)x1)



and limx+ln(x)x=0



thus



lim+(ln(x)x)=


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