Sunday, 8 February 2015

calculus - is it for granted that logx grows slower than xa?




When facing this problem limxlogxxa

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=11txt=1tc1=xc1c<xcc


for all x>1; if c:=a/2, then
logxxa<xcac=2axa/20

as x grows indefinitely.



Now you may extend the argument above to an epsilon-argument easily.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...