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$f:(0,\infty)\rightarrow \Bbb R$, $x\mapsto 1/x$

show that $f$ is continuous.

To prove the continuity on the given domain, for each $x_0\in(0,\infty)$ and $\epsilon>0$ we need to determine a $\delta(x_0,\epsilon)$ which satisfies $|x-x_0|<\delta(x_0,\epsilon)\Rightarrow|f(x)-f(x_0)|\lt \epsilon$.

In which way one could easily find $\delta(x_0,\epsilon)$?

M. Winter
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delog
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    You mixed up the quantifiers badly, the $\delta$ can (and must) depend on $e$ (or better $\epsilon$) and $x_0$. You will not find a $\delta$ such that this statement is true for all $e$. – M. Winter Jun 07 '17 at 10:54
  • Finding a $\delta$ that works at every point in the domain means that $f$ is uniformly continuous - a stronger notion. – Prahlad Vaidyanathan Jun 07 '17 at 10:57
  • @M.Winter would you please edit the OP? I didn't get it clearly because of my unfamiliarity over quantifiers – delog Jun 07 '17 at 11:04
  • I suggest starting by writing $|f(x)-f(x_0)|$ explicitly as a fraction. Then upper bound this fraction by using $x\geq x_0-\delta$. – Mau314 Jun 07 '17 at 11:22

1 Answers1

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Let $x_0\in (0,+\infty) $.

For $x\in (\frac {x_0}{2},\frac {3x_0}{2} ), $ we have $$|x-x_0|<\frac {x_0}{2} \tag 1$$ and $$|\frac {1}{x}-\frac {1}{x_0}|=\frac{|x-x_0|}{xx_0}$$ $$<2\frac {|x-x_0|}{x_0^2} \tag 2$$

thus, to satisfy the condition $$|\frac {1}{x}-\frac {1}{x_0}|<\epsilon $$ it is sufficient to have

$|x-x_0|<\frac {x_0}{2}$ and $|x-x_0|<x_0^2\frac {\epsilon}{2} $

from this, we can take $$\delta=\min (\frac {x_0}{2},x_0^2\frac {\epsilon}{2}) .$$

hamam_Abdallah
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