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If-f-x-x-2-4-x-2-x-2-3-0-1-find-Alpha-3-Epsalum-0-1-Find-delta-

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Question Number 5232 by sanusihammed last updated on 02/May/16
If f(x) = ((x^2 −4)/(x−2)) x→2 α = 3 . Σ = 0.1 find δ Alpha = 3 Epsalum = 0.1 Find delta
$${If}\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}−\mathrm{2}}\:\:\:\: \\ $$$$\:\:\:\:{x}\rightarrow\mathrm{2} \\ $$$$ \\ $$$$\alpha\:=\:\mathrm{3}\:.\:\Sigma\:=\:\mathrm{0}.\mathrm{1}\:\:\:{find}\:\:\:\delta \\ $$$$ \\ $$$${Alpha}\:=\:\mathrm{3}\:\:\:{Epsalum}\:=\:\mathrm{0}.\mathrm{1}\:{Find}\:{delta} \\ $$
Commented by FilupSmith last updated on 02/May/16
What is the definition of δ?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\delta? \\ $$
Answered by Yozzii last updated on 02/May/16
According to the ε−δ definition of the limit, the limit lim_(x→a) f(x)=α exists if,for any ε>0 there is a δ>0, such that ∣f(x)−α∣<ε whenever 0<∣x−a∣<δ. So using f(x)=((x^2 −4)/(x−2)) and α=4 {not 3 since lim_(x→2) f(x)=lim_(x→2) (((x+2)(x−2))/(x−2))=lim_(x→2) (x+2)=2+2=4≠3.} ⇒∣((x^2 −4)/(x−2))−4∣=∣((x^2 −4−4x+8)/(x−2))∣ =∣((x^2 −4x+4)/(x−2))∣=∣(((x−2)^2 )/(x−2))∣ =∣x−2∣ Since ∣f(x)−4∣<ε and ∣f(x)−4∣=∣x−2∣ ⇒∣x−2∣<ε⇒ let ε=δ. Hence, if ε=0.1, δ=0.1.
$${According}\:{to}\:{the}\:\epsilon−\delta\:{definition}\:{of} \\ $$$${the}\:{limit},\:{the}\:{limit}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\alpha \\ $$$${exists}\:{if},{for}\:{any}\:\epsilon>\mathrm{0}\:{there}\:{is}\:{a}\:\delta>\mathrm{0}, \\ $$$${such}\:{that}\: \\ $$$$\mid{f}\left({x}\right)−\alpha\mid<\epsilon\:\:{whenever}\:\:\mathrm{0}<\mid{x}−{a}\mid<\delta. \\ $$$${So}\:{using}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}−\mathrm{2}}\:{and}\:\alpha=\mathrm{4}\:\:\:\:\left\{{not}\:\mathrm{3}\:{since}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}{{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left({x}+\mathrm{2}\right)=\mathrm{2}+\mathrm{2}=\mathrm{4}\neq\mathrm{3}.\right\} \\ $$$$\Rightarrow\mid\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}−\mathrm{2}}−\mathrm{4}\mid=\mid\frac{{x}^{\mathrm{2}} −\mathrm{4}−\mathrm{4}{x}+\mathrm{8}}{{x}−\mathrm{2}}\mid \\ $$$$=\mid\frac{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}{{x}−\mathrm{2}}\mid=\mid\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{{x}−\mathrm{2}}\mid \\ $$$$=\mid{x}−\mathrm{2}\mid \\ $$$${Since}\:\mid{f}\left({x}\right)−\mathrm{4}\mid<\epsilon\:{and}\:\mid{f}\left({x}\right)−\mathrm{4}\mid=\mid{x}−\mathrm{2}\mid \\ $$$$\Rightarrow\mid{x}−\mathrm{2}\mid<\epsilon\Rightarrow\:{let}\:\epsilon=\delta.\:{Hence},\:{if} \\ $$$$\epsilon=\mathrm{0}.\mathrm{1},\:\delta=\mathrm{0}.\mathrm{1}. \\ $$

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