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I want to proof that the function $\frac{1}{1+x}$ is continuous in the positive real numbers. Therefore I use the epsilon delta definition of continuity. But when I make my calculations (to find epsilon in function of delta) I get stuck.

Can anybody help me to go further on $\left| \frac{x-a}{(1-x)(1-a)} \right| < \epsilon$.

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  • $\begingroup$ Why do you have $(1-x)(1-a)$ in the denominator? $\endgroup$ Feb 25, 2021 at 19:12
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    $\begingroup$ Because according to the defintion |f(x)-f(a)| u have $| \frac{1}{1-x} - \frac{1}{1-a}|$ and then make them have the same denominator. $\endgroup$ Feb 25, 2021 at 19:24
  • $\begingroup$ except the function is $\frac1{1+x},$ not $\frac1{1-x}.$ $\endgroup$ Feb 25, 2021 at 19:26
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    $\begingroup$ Ow sorry quick mistake! $\endgroup$ Feb 25, 2021 at 19:30

1 Answer 1

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It should read: $\left|\dfrac{x-a}{(1+x)(1+a)}\right| < |x-a| < \epsilon$ if you let $\delta = \epsilon > 0$.

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    $\begingroup$ I see. This is because |1/(1+x)| is less than one for every x of the positive real numbers? $\endgroup$ Feb 25, 2021 at 19:09
  • $\begingroup$ @notaMathqueen: Yes it is ! $\endgroup$
    – user781811
    Feb 25, 2021 at 19:09

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