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Question Number 4728 by 123456 last updated on 01/Mar/16

if ∣x^2 −c∣≤ε, c∈R,c≥0 does? ∣x−(√c)∣≤ε

if∣x2−c∣⩽ϵ,c∈R,c⩾0does?∣x−c∣⩽ϵ

Commented by Yozzii last updated on 01/Mar/16

∣(x−(√c))∣∣x+(√c)∣≤ε ∣x−(√c)∣≤(ε/(∣x+(√c)∣)) (x≠−(√c)) If ∣x+(√c)∣≥1⇒1≥(1/(∣x+(√c)∣)) ⇒ε≥(ε/(∣x+(√c)∣))⇒∣x−(√c)∣≤ε. It appears to be necessary that x≥1−(√c) or x≤−1−(√c) to deduce ∣x−(√c)∣≤ε (ε>0). Suppose x=1−(√(25))−1=−5 for c=25 (so∣x−(√c)∣<1). ⇒∣25−25∣=0≤1 (so say ε=1). But∣x−(√c)∣=∣−5−5∣=101=ε. So, (x,c,ε)=(−5,25,1) shows the original implication being incorrect.

∣(x−c)∣∣x+c∣⩽ϵ∣x−c∣⩽ϵ∣x+c∣(x≠−c)If∣x+c∣⩾1⇒1⩾1∣x+c∣⇒ϵ⩾ϵ∣x+c∣⇒∣x−c∣⩽ϵ.Itappearstobenecessarythatx⩾1−corx⩽−1−ctodeduce∣x−c∣⩽ϵ(ϵ>0).Supposex=1−25−1=−5forc=25(so∣x−c∣<1).⇒∣25−25∣=0⩽1(sosayϵ=1).But∣x−c∣=∣−5−5∣=10≰1=ϵ.So,(x,c,ϵ)=(−5,25,1)showstheoriginalimplicationbeingincorrect.

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