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let-f-0-R-be-a-continuous-function-if-lim-n-0-1-f-x-n-dx-2-then-lim-n-f-nx-




Question Number 206433 by universe last updated on 14/Apr/24
     let f:[0,∞)→R be a continuous function if      lim_(n→∞ )  ∫_0 ^1 f(x+n)dx = 2   then lim_(n→∞)  f(nx) = ?
letf:[0,)Rbeacontinuousfunctioniflimn01f(x+n)dx=2thenlimnf(nx)=?
Answered by Berbere last updated on 14/Apr/24
∫_0 ^1 f(x+n)dx=^(x+n=y) ∫_n ^(1+n) f(y)dy=2  if lim_(x→∞) f(x) existe let a=lim_(x→∞) f(x)  ⇒∀ε>0 ∃N such that ∀ x>N  a−ε≤f(x)≤a+ε  ⇒a−ε≤∫_n ^(n+1) f(x)≤a+ε;∀n>N  ⇒lim_(n→∞) ∫_n ^(n+1) f(x)dx=a=2  then lim_(n→∞) f(nx)= { ((f(0) if x=0)),((2 ∀x>0)) :}
01f(x+n)dx=x+n=yn1+nf(y)dy=2iflimxf(x)existeleta=limxf(x)ϵ>0Nsuchthatx>Naϵf(x)a+ϵaϵnn+1f(x)a+ϵ;n>Nlimnnn+1f(x)dx=a=2thenlimnf(nx)={f(0)ifx=02x>0
Commented by aleks041103 last updated on 14/Apr/24
f is continious ⇏ ∃lim_(x→∞)  f(x)  example: f(x)=sin(x), ∄lim f(x)_(x→∞)   therefore your solution is wrong
fiscontiniouslimxf(x)example:f(x)=sin(x),limf(x)xthereforeyoursolutioniswrong
Commented by Berbere last updated on 14/Apr/24
yes you right i dont why i said That
yesyourightidontwhyisaidThat
Answered by Berbere last updated on 14/Apr/24
f(x)=2+cos(2πx)  ∫_0 ^1 (2+cos(2πx))=2 no limite
f(x)=2+cos(2πx)01(2+cos(2πx))=2nolimite

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