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Question Number 1499 by 112358 last updated on 14/Aug/15

Use the ε−δ definition of the limit to show that lim_(x→3) (x/(6−x))=1 . (I′m hoping to better understand this concept by example so please help me by explaning the reasoning behind your steps.)

$${Use}\:{the}\:\epsilon−\delta\:{definition}\:{of}\:{the} \\ $$$${limit}\:{to}\:{show}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{6}−{x}}=\mathrm{1}\:. \\ $$$$\left({I}'{m}\:{hoping}\:{to}\:{better}\:{understand}\right. \\ $$$${this}\:{concept}\:{by}\:{example}\:{so}\:{please} \\ $$$${help}\:{me}\:{by}\:{explaning}\:{the}\: \\ $$$$\left.{reasoning}\:{behind}\:{your}\:{steps}.\right) \\ $$

Commented by 123456 last updated on 14/Aug/15

∀ε>0,∃δ,∀x,0<∣x−3∣<δ,∣(x/(6−x))−1∣<ε

$$\forall\epsilon>\mathrm{0},\exists\delta,\forall{x},\mathrm{0}<\mid{x}−\mathrm{3}\mid<\delta,\mid\frac{{x}}{\mathrm{6}−{x}}−\mathrm{1}\mid<\epsilon \\ $$

Commented by 123456 last updated on 14/Aug/15

(x/(6−x))−1=((x−(6−x))/(6−x))=((2x−6)/(6−x))=((2(x−3))/(6−x))

$$\frac{{x}}{\mathrm{6}−{x}}−\mathrm{1}=\frac{{x}−\left(\mathrm{6}−{x}\right)}{\mathrm{6}−{x}}=\frac{\mathrm{2}{x}−\mathrm{6}}{\mathrm{6}−{x}}=\frac{\mathrm{2}\left({x}−\mathrm{3}\right)}{\mathrm{6}−{x}} \\ $$

Commented by 123456 last updated on 14/Aug/15

(x/(6−x))=k⇔x=6k−kx⇔x=((6k)/(1+k)) k=1+ε⇒x=((6(1+ε))/(2+ε))

$$\frac{{x}}{\mathrm{6}−{x}}={k}\Leftrightarrow{x}=\mathrm{6}{k}−{kx}\Leftrightarrow{x}=\frac{\mathrm{6}{k}}{\mathrm{1}+{k}} \\ $$$${k}=\mathrm{1}+\epsilon\Rightarrow{x}=\frac{\mathrm{6}\left(\mathrm{1}+\epsilon\right)}{\mathrm{2}+\epsilon} \\ $$

Answered by 123456 last updated on 16/Aug/15

lim_(x→3) (x/(6−x))=1⇒∃ε>0,∀δ,0<∣x−3∣<δ⇒∣(x/(6−x))−1∣<ε lets simplyfy things a little ∣(x/(6−x))−1∣=∣((x−(6−x))/(6−x))∣=∣((x−6+x)/(6−x))∣=∣((2x−6)/(6−x))∣ =((2∣x−3∣)/(∣6−x∣))<ε ∣x−3∣<((∣6−x∣ε)/2) then ∣x−3∣<1⇒2<x<4⇒−4<x<−2⇒2<6−x<4⇒2<∣6−x∣<4 ∣x−3∣<((∣6−x∣ε)/2)<((2ε)/2)=ε then lets δ=min(1,ε) if δ=1, then 1<ε⇒ε>1 ∣x−3∣<1⇒((2∣x−3∣)/(∣6−x∣))<(2/(∣6−x∣)) 2<x<4⇒2<6−x<4⇒(1/4)<(1/(6−x))<(1/2)⇒(1/4)<(1/(∣6−x∣))<(1/2) ((2∣x−3∣)/(∣6−x∣))<(2/(∣6−x∣))<(2/2)<1<ε ∣(x/(6−x))−1∣<ε then suppose that δ=ε ∣x−3∣<δ⇒((2∣x−3∣)/(∣6−x∣))<((2ε)/2)=ε ∣(x/(6−x))−1∣<ε

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{{x}}{\mathrm{6}−{x}}=\mathrm{1}\Rightarrow\exists\epsilon>\mathrm{0},\forall\delta,\mathrm{0}<\mid{x}−\mathrm{3}\mid<\delta\Rightarrow\mid\frac{{x}}{\mathrm{6}−{x}}−\mathrm{1}\mid<\epsilon \\ $$$$\mathrm{lets}\:\mathrm{simplyfy}\:\mathrm{things}\:\mathrm{a}\:\mathrm{little} \\ $$$$\mid\frac{{x}}{\mathrm{6}−{x}}−\mathrm{1}\mid=\mid\frac{{x}−\left(\mathrm{6}−{x}\right)}{\mathrm{6}−{x}}\mid=\mid\frac{{x}−\mathrm{6}+{x}}{\mathrm{6}−{x}}\mid=\mid\frac{\mathrm{2}{x}−\mathrm{6}}{\mathrm{6}−{x}}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\mid{x}−\mathrm{3}\mid}{\mid\mathrm{6}−{x}\mid}<\epsilon \\ $$$$\mid{x}−\mathrm{3}\mid<\frac{\mid\mathrm{6}−{x}\mid\epsilon}{\mathrm{2}} \\ $$$$\mathrm{then} \\ $$$$\mid{x}−\mathrm{3}\mid<\mathrm{1}\Rightarrow\mathrm{2}<{x}<\mathrm{4}\Rightarrow−\mathrm{4}<{x}<−\mathrm{2}\Rightarrow\mathrm{2}<\mathrm{6}−{x}<\mathrm{4}\Rightarrow\mathrm{2}<\mid\mathrm{6}−{x}\mid<\mathrm{4} \\ $$$$\mid{x}−\mathrm{3}\mid<\frac{\mid\mathrm{6}−{x}\mid\epsilon}{\mathrm{2}}<\frac{\mathrm{2}\epsilon}{\mathrm{2}}=\epsilon \\ $$$$\mathrm{then}\:\mathrm{lets} \\ $$$$\delta=\mathrm{min}\left(\mathrm{1},\epsilon\right) \\ $$$$\mathrm{if}\:\delta=\mathrm{1},\:\mathrm{then}\:\mathrm{1}<\epsilon\Rightarrow\epsilon>\mathrm{1} \\ $$$$\mid{x}−\mathrm{3}\mid<\mathrm{1}\Rightarrow\frac{\mathrm{2}\mid{x}−\mathrm{3}\mid}{\mid\mathrm{6}−{x}\mid}<\frac{\mathrm{2}}{\mid\mathrm{6}−{x}\mid} \\ $$$$\mathrm{2}<{x}<\mathrm{4}\Rightarrow\mathrm{2}<\mathrm{6}−{x}<\mathrm{4}\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}<\frac{\mathrm{1}}{\mathrm{6}−{x}}<\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}<\frac{\mathrm{1}}{\mid\mathrm{6}−{x}\mid}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}\mid{x}−\mathrm{3}\mid}{\mid\mathrm{6}−{x}\mid}<\frac{\mathrm{2}}{\mid\mathrm{6}−{x}\mid}<\frac{\mathrm{2}}{\mathrm{2}}<\mathrm{1}<\epsilon \\ $$$$\mid\frac{{x}}{\mathrm{6}−{x}}−\mathrm{1}\mid<\epsilon \\ $$$$\mathrm{then}\:\mathrm{suppose}\:\mathrm{that}\:\delta=\epsilon \\ $$$$\mid{x}−\mathrm{3}\mid<\delta\Rightarrow\frac{\mathrm{2}\mid{x}−\mathrm{3}\mid}{\mid\mathrm{6}−{x}\mid}<\frac{\mathrm{2}\epsilon}{\mathrm{2}}=\epsilon \\ $$$$\mid\frac{{x}}{\mathrm{6}−{x}}−\mathrm{1}\mid<\epsilon \\ $$

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