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Question Number 4297 by 123456 last updated on 07/Jan/16
lets  f:[0,+∞)→R,∀x≥y⇒f(x)≥f(y)  g:[0,+∞)→R  if  ∀x∈[0,+∞),f(x)≤g(x)≤f(2x)  lim_(x→+∞) f(x)=L,L is finite  does  lim_(x→+∞) f(x)−g(x)=0?
letsf:[0,+)R,xyf(x)f(y)g:[0,+)Rifx[0,+),f(x)g(x)f(2x)limx+f(x)=L,Lisfinitedoeslimx+f(x)g(x)=0?
Commented by Yozzii last updated on 08/Jan/16
lim_(x→∞) f(2x)=lim_(0.5u→∞) f(u)=lim_(u→∞) f(u)=L  Since f(x)≤g(x)≤f(2x) ∀x∈[0,∞)  ⇒lim_(x→∞) f(x)≤lim_(x→∞) g(x)≤lim_(x→∞) f(2x).  lim_(x→∞) f(x)=lim_(x→∞) f(2x)=L ⇒lim_(x→∞) g(x)=L  by the squeeze theorem.  lim_(x→∞) {f(x)−g(x)}=lim_(x→∞) f(x)−lim_(x→∞) g(x)                                    =L−L                                    =0.  (Under construction)
limxf(2x)=lim0.5uf(u)=limuf(u)=LSincef(x)g(x)f(2x)x[0,)limxf(x)limxg(x)limxf(2x).limxf(x)=limxf(2x)=Llimxg(x)=Lbythesqueezetheorem.limx{f(x)g(x)}=limxf(x)limxg(x)=LL=0.(Underconstruction)
Commented by prakash jain last updated on 08/Jan/16
f(x)=x  g(x)=1.5x  x≥0, f(x)≤g(x)≤f(2x)  f(x)−g(x)=−.5x  lim_(x→+∞) f(x)−g(x)≠0
f(x)=xg(x)=1.5xx0,f(x)g(x)f(2x)f(x)g(x)=.5xlimx+f(x)g(x)0
Commented by Yozzii last updated on 08/Jan/16
lim_(x→∞) f(x)=lim_(x→∞)  x=∞ (not finite)  I′m searching for a counterexample  as well.
limxf(x)=limxx=(notfinite)Imsearchingforacounterexampleaswell.
Commented by prakash jain last updated on 08/Jan/16
oh. I did not pay attention.
oh.Ididnotpayattention.
Commented by prakash jain last updated on 08/Jan/16
You are correct.   lim_(x→∞) f(x)−g(x)=0
Youarecorrect.limxf(x)g(x)=0

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