When facing this problem limx→∞logxxa
with a>0, then I automatically think of l'Hoital's. But then can we approach this problem using the definition of limits? I mean can we find x0 such that for x>x0(ϵ) it holds that logxxa<ϵ for any ϵ>0?
Essentially is it for granted that logx grows slower than xa? What is the simplest thing we could assume to have this proved?
Answer
Let a>0; note that for all c>0 we have
logx=∫xt=11t≤∫xt=1tc−1=xc−1c<xcc
for all x>1; if c:=a/2, then
logxxa<xc−ac=2ax−a/2→0
as x grows indefinitely.
Now you may extend the argument above to an epsilon-argument easily.
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