Prove that:



Basically need to prove that : $\forall\varepsilon>0\,\,\exists\delta>0\forall x\in\mathbb{R}\quad-\delta<x-11<0\rightarrow\left|\sqrt{x-2}-3\right|<\varepsilon$

Perpwork: Let $\varepsilon>0$ be given. $$\begin{array}{cc} & \left|\sqrt{x-2}-3\right|<\varepsilon\\ \iff & -\varepsilon<\sqrt{x-2}-3<\varepsilon\\ \iff & 3-\varepsilon<\sqrt{x-2}<3+\varepsilon\\ \iff & \left(3-\varepsilon\right)^{2}<|x-2|<\left(3+\varepsilon\right)^{2}\\ \\ \end{array}$$

Given that $ \forall\varepsilon>0\quad\left(3-\varepsilon\right)^{2}\geq0$

$$\begin{array}{cc} & \left(3-\varepsilon\right)^{2}<|x-2|<\left(3+\varepsilon\right)^{2}\\ \rightarrow & \left(3-\varepsilon\right)^{2}<x-2<\left(3+\varepsilon\right)^{2}\\ \iff & \left(3-\varepsilon\right)^{2}+2<x<\left(3+\varepsilon\right)^{2}+2\\ \iff & \left(3-\varepsilon\right)^{2}-9<x-11<\left(3+\varepsilon\right)^{2}-9 \end{array}$$

Set $\delta<9-\left(3-\varepsilon\right)^{2}$

$$\begin{array}{ccccc} & -\delta<x-11<0\\ \iff & \left(3-\varepsilon\right)^{2}-9<x-11 & \rightarrow & \left(3-\varepsilon\right)^{2}<|x-2|<\left(3+\varepsilon\right)^{2}\\ & & \rightarrow & -\varepsilon<\sqrt{x-2}-3<\varepsilon\\ & & \rightarrow & \left|\sqrt{x-2}-3\right|<\varepsilon \end{array}$$ Q.E.D

Is this a legit proof ?

  • $\begingroup$ Thanks alex for your reply. I used the rule that every real number squared is non negative the line before. Is it something that doesn't counts there? $\endgroup$ – Pavel Penshin Apr 6 '16 at 17:39
  • $\begingroup$ Apologies, in your case it is. What I mean that for example if $0.1 <|x-2|<7$ then it doesn't follow that $0.1<x-2<7$ as $x-2$ can be negative $\endgroup$ – Alex Apr 6 '16 at 20:45
  • $\begingroup$ It opens up into two options the other one isn't in the real set ... $\endgroup$ – Pavel Penshin Apr 6 '16 at 20:48

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