Friday, 12 August 2016

real analysis - Finding limxtoinftyleft(fraca1/x1+a1/x2+cdotcdotcdotan1/xnright)nx



For non zero positive reals a1,a2an how to find
limx(a1/x1+a1/x2++a1/xnn)nx?




It becomes indeterminant form 1. But difficult to solve by L'Hospital's Rule. By using A.M.-G.M. inequality it comes that limit is a1a2an. I also tried by using Squeeze. Please help.Thanks.


Answer



tl;dr




For a1,,an positive numbers, and x>0, limx(a1x1+a1x2++an1xn)nx=nk=1ak





Proof. Detailed approach, where we use the (low-order) Taylor expansions eu=1+u+o(u) and ln(1+u)=u+o(u) when u0.



You can rewrite
(a1x1+a1x2++an1xn)nx=exp(nxlna1x1+a1x2++an1xn)=exp(nxln1nnk=1a1xk)=exp(nxln1nnk=1e1xlnak)


Since n is a constant and 1xlnakx0 for each k, we have
e1xlnak=1+1xlnak+o(1x)

for each k as x, and therefore
(a1x1+a1x2++an1xn)nx=exp(nxln1nnk=1e1xlnak)=exp(nxln(1+1nxnk=1lnak+o(1x)))=exp(nx(1nxnk=1lnak+o(1x)))=exp(nk=1lnak+o(1))=exp(lnnk=1ak+o(1))=eo(1)nk=1akxnk=1ak.


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