Sunday, September 25, 2016

Calculating integral without L'hopital's rule




Problemlimx01xx0(1+u)1udu=?





My first solution is using L'hopital's rule




Solution using L'hopital's rule, given limit
limx0(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)>(1ϵ)u for $0. Hence e(1ϵ)(1+u)1/ue for $0 provided $0. Squeeze theorem completes the proof since ϵ is arbitrary.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...