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I need to prove that the function $f(x)=(x+1)/(x-1)$ is continuous on the interval $(1,∞)$ using the epsilon-delta definition of continuity.

Here's my work thus far:

Let $x,y\in(1,∞)$. Then $$|f(x)-f(y)|=\left|\frac{x+1}{x-1}-\frac{y+1}{y-1}\right|\\ =\left|\frac{-2x+2y}{(x-1)(y-1)}\right|\\ =2\frac{|x-y|}{|x-1||y-1|}\\ =2\frac{|x-y|}{(x-1)(y-1)}$$

Then, if we have $|x-y|<\delta$ where $\delta>0$, we get $2\frac{|x-y|}{(x-1)(y-1)}<2\frac{\delta}{(x-1)(y-1)}$.

I don't know how to proceed from here, since I can't really make (x-1) and (y-1) any smaller in order to get $|f(x)-f(y)|<\epsilon$ which I need to prove that the function is continuous. Any help is appreciated!

(sorry if this makes no sense, I am not a native English speaker...)

Arthur
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pileafilea
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    You're almost there. What happens to $1/(x-1)$ as $x$ gets larger? – Aaron Goldsmith Nov 14 '22 at 21:07
  • Have you seen a proof of the fact that $f(x)/g(x)$ is continuous on any interval on which $f$ and $g$ are both continuous and $g$ is non-zero? If so then replay the proof for this specific example of $f$ and $g$. – Rob Arthan Nov 14 '22 at 23:11
  • Do you mean$ f:(1,\infty) \rightarrow \mathbb R)$ with the usual topology on $\mathbb R$ and the subset topology on $(1,\infty)$ induced by the usual topology? Questions about continuity do not make sense unless you specify the topologies on the domain and co-domain. – P. Lawrence Nov 14 '22 at 23:18
  • @P.Lawrence if we are doing real analysis it is always assumed the usual topology unless otherwise specified. – Marcos Nov 14 '22 at 23:23

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It's better to replace $y$ with $x_0$ for comprehension that we can fix it and let $x$ vary.

In fact you want to prove continuity in $x_0\in[1+c,\infty)$ for arbitrary $c$.

So $x_0\ge 1+c$ and $|x-x_0|<\delta\implies x\ge 1+c-\delta$

Reporting in your calculation we get $|f(x)-f(x_0)|\le \dfrac{2\delta}{c(c-\delta)}$

For this quantity to be less than epsilon, you need $\delta<\dfrac{c^2\epsilon}{2+c\epsilon}$

Note that the RHS is always $<c$ so $x>1$ stays satisfied.

zwim
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