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Proving $\frac{x}{x-1}$ is uniformly continuous on the interval $(2,4)$.

By definition, $\lvert\frac{x}{x-1}- \frac{y}{y-1}\rvert< \epsilon$. As $4 >x,y > 2$ we can state: $\lvert\frac{x}{x-1} - \frac{y}{y-1}\rvert< \lvert\frac{x}{4-1} -\frac{ y}{4-1}\rvert < 1/3|x-y| < \epsilon/3$. Therefore $\delta = \min\{4,\epsilon/3\}.$

Is this the right approach? How do I prove uniform continuity on the upper bound $4$?

Any hints/help is appreciated.

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  • $\begingroup$ You can show the function is continuous in $[2,4]$ and use Cantor's theorem. $\endgroup$ – Itay4 Apr 26 '17 at 11:56
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Apr 26 '17 at 11:57
  • $\begingroup$ No...it seems problematic for (i) You shall pick $\epsilon$ first then show that there is a $\delta$ such that.... (ii) $x, y \in (2, 4)$ does not imply $ |x/(x -1) - y/(y - 1)| < |x - y|/3$. Take $x = 2.1, y = 2.4$, then the left hand side is 0.1948, the right hand side is 0.1. $\endgroup$ – Li Chun Min Apr 26 '17 at 12:22
  • $\begingroup$ Where should I start to prove this then? :S $\endgroup$ – Pixel Rain Apr 26 '17 at 12:23
  • $\begingroup$ See my answer. Let me have my dinner first:) $\endgroup$ – Li Chun Min Apr 26 '17 at 12:33
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Pick any $\epsilon > 0$. Take $\delta = \epsilon$. Pick any $x, y \in (2,4)$ such that $|x - y| < \delta$.

$$\bigg| \frac{x}{x -1} - \frac{y}{y-1}\bigg| =\bigg| \frac{x-y}{(x -1)(y -1)} \bigg| = \frac{|x - y|}{|x - 1||y -1|} < \frac{\delta}{|x - 1||y -1|} $$ But we have that $x - 1 > 1$ and $y -1> 1$. So that $$\bigg| \frac{x}{x -1} - \frac{y}{y-1}\bigg| < \frac{\delta}{|x - 1||y -1|} < \delta. $$

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  • $\begingroup$ if (x-1) > 1, why is that the inequality is < delta? Wouldn't the fraction with |x-1||y-1| > 1? $\endgroup$ – Pixel Rain Apr 26 '17 at 12:40
  • $\begingroup$ Then $|x -1|>1$ and taking reciprocal we get $1/|x -1| < 1$. $\endgroup$ – Li Chun Min Apr 26 '17 at 12:42
  • $\begingroup$ Ahhh that makes sense, thank you! $\endgroup$ – Pixel Rain Apr 26 '17 at 12:44
  • $\begingroup$ you're welcome! $\endgroup$ – Li Chun Min Apr 26 '17 at 12:44
  • $\begingroup$ @PixelRain It is a good exercise to prove that your function is (i) continuous in $(1, +\infty)$. (ii) NOT uniformly continuous on $(1, +\infty)$. $\endgroup$ – Li Chun Min Apr 26 '17 at 12:51

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