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Question Number 72806 by Rio Michael last updated on 03/Nov/19

 show that  lim_(                                                   x→0)  [ x]  does not exist.  Hence define  [x]  and sketch a graph for    y = 3x^2  + [x]

showthatlimx0[x]doesnotexist.Hencedefine[x]andsketchagraphfory=3x2+[x]

Commented by mathmax by abdo last updated on 03/Nov/19

let f(x)=[x] we have lim_(x→0^+ )   f(x)=lim_(x→0^+ )   [x] =0  lim_(x→0^− )   f(x)=lim_(x→0^− )    [x]=−1  and f(0)=0 we have 0≠−1 ⇒  lim_(x→0) [x] dont exist

letf(x)=[x]wehavelimx0+f(x)=limx0+[x]=0limx0f(x)=limx0[x]=1andf(0)=0wehave01limx0[x]dontexist

Commented by Rio Michael last updated on 03/Nov/19

love that method, thanks sir

lovethatmethod,thankssir

Answered by mind is power last updated on 03/Nov/19

suppose that lim_(x→0)  [x] exist  call l=lim_(x→0) [x]  ⇒∀ε>0 ∃η>0∣   ∣x∣<η⇒∣[x]−l∣<ε  let ε=(l/3)  ⇒∃η_ε  such that ∀x∈]−η,η[  ∣[x]−L∣<(1/3)  for x∈]−η,0[ ⇒∣−1−L∣<(1/2)⇔∣1+L∣<(1/3)  for x∈]0,η[⇒∣−L∣<(1/3)  we have ∣1+L−L∣=1  truangular inquality⇒∣1∣=1≤∣−L∣+∣1+L∣≤(1/3)+(1/3)=(2/3)  absurd since that L/dosent exist    2)   y=3x^2 +k  x∈[k,k+1[

supposethatlimx0[x]existcalll=limx0[x]ε>0η>0x∣<η⇒∣[x]l∣<εletϵ=l3ηεsuchthatx]η,η[[x]L∣<13forx]η,0[⇒∣1L∣<12⇔∣1+L∣<13forx]0,η[⇒∣L∣<13wehave1+LL∣=1truangularinquality⇒∣1∣=1⩽∣L+1+L∣⩽13+13=23absurdsincethatL/dosentexist2)y=3x2+kx[k,k+1[

Commented by Rio Michael last updated on 03/Nov/19

thanks

thanks

Commented by mind is power last updated on 03/Nov/19

most Welcom

mostWelcom

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