Problemlimx→01x∫x0(1+u)1udu=?
My first solution is using L'hopital's rule
Solution using L'hopital's rule, given limit
limx→0(1+x)1x1=e
I want another solution without using l'hopital's rule, but I can't found it. More annoying thing is I can't even integrate ∫x0(1+u)1udu. Or It can't be solved without l'hopital's rule?
Answer
(1+u)1/u=e1ulog(1+u)≤e. Also log(1+u)=u+o(u) so, given ϵ>0 there exists δ such that log(1+u)>(1−ϵ)u for $0. Hence e(1−ϵ)≤(1+u)1/u≤e for $0 provided $0
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