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Please help me to prove that the functions $f(x)=x$ and $g(x)=x−\frac{1}{x}$ satisfy the following property: $$ \forall\varepsilon>0\quad\exists\delta>0:\quad\forall x>0\quad(|f(x)-g(x)|<\delta\Rightarrow f(x)g(x)>\varepsilon) $$

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What have you tried? Take any $\epsilon>0$. You have $|f(x)-g(x)|=1/|x|<\delta$, where you now want to pick a $\delta$ so that $f(x)g(x)>\epsilon$... – Alex R. Sep 15 '12 at 17:30

You have $$f(x)g(x)=x^2-1>\epsilon$$ which leads you; $x^2>1+\epsilon$ and therefore $|x|>\sqrt{1+\epsilon}$. From $|f(x)-g(x)|<\delta$ you get $|x|>1/\delta$. Now Taking $\delta=\large \frac{1}{\sqrt{1+\epsilon}}$ for all positive $\epsilon$ makes your statement correct. In fact by choosing $\epsilon>0$ and putting $\delta=\large \frac{1}{\sqrt{1+\epsilon}}$ you are writing a logical statement.

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$|f(x)-g(x)|<\delta$ iff $|x|>\frac{1}{\delta}$, and if $|x|$ is large enough, $f(x)g(x)\approx x^2$, which can be made as large as you want.

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