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Find $\sup{A},\inf{A},\max{A},\min{A}$ where: $$A=\left\lbrace\frac{2013}{1+\epsilon+\epsilon^{-1}}:\epsilon\in(0,1)\right\rbrace$$

I suspect that $\sup{A}=\frac{2013}{3}, \inf{A}=0$ and max and min don't exist, I can easily prove that my candidates are upper and lower bounds, but how to proceed from there?

Jimmy R.
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qiubit
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1 Answers1

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Your answer is correct. You can prove it by using that $A$ is strictly increasing in $\epsilon$, in the interval $(0,1)$ since $$\frac{∂A}{∂\epsilon}=−2013\frac{\epsilon^2-1}{\epsilon^4+2\epsilon^3+3\epsilon^2+2\epsilon+1}>0, \qquad \forall\epsilon\in(0,1)$$

Jimmy R.
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  • I'm not allowed to use derivatives. Also why can I say that $\frac{2013}{3}$ is the sup I'm looking for knowing only that the function is strictly increasing on a given interval? – qiubit Nov 06 '14 at 16:27
  • @qiubit: What are you allowed to use? – Jonas Meyer Nov 06 '14 at 16:31
  • Only the definition of sup and inf. I can also use limits but only for sequences not functions – qiubit Nov 06 '14 at 16:32