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Given, $g(x) = \frac{1}{1-x} + 1$. I want to prove that $g$ is continuous at $x=0$. I specifically want to do an $\epsilon-\delta$ proof. Related to this Is $g(x)\equiv f(x,1) = \frac{1}{1-x}+1$ increasing or decreasing? differentiable $x=1$?.

My work: Let $\epsilon >0$ be given. The challenging part for me is to pick the $\delta$.

Let $\delta = 2 \epsilon$ and suppose that $|x-0| = |x| < \delta$ and $|\frac{1}{1-x}| < \frac{1}{2}$.

So $$|g(x) - g(0)|$$ $$=|\frac{2-x}{1-x} - 2|$$ $$=|\frac{x}{1-x}|$$ $$=|\frac{1}{1-x}||x|$$ $$<\frac{1}{2} 2 \epsilon = \epsilon$$.

So $g$ is continuous at $x=0$. Is my proof correct?

EDIT: Rough work on how I picked $\delta$. Suppose $|x-0|<\delta$ and since $\delta \leq 1$, we have $|x|<1$, so $-1<x<1$ Then this implies $0<1-x<2$. The next step I'm not too sure of, I have $0<\frac{1}{1-x}<\frac{1}{2} \implies |\frac{1}{1-x}|<\frac{1}{2}$.

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OGC
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2 Answers2

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Hint. See my comment to your post.

After doing your algebra, as you have shown, $$\left|\dfrac{1}{1-x}+1-2\right| = \left|x\right|\left|\dfrac{1}{1-x}\right|\text{.}$$ We have $|x| < \delta$. Let $\delta = 1/2$, then $$|x| < \delta \implies -\delta < x < \delta \implies 1-\delta < 1-x < 1 + \delta \implies \dfrac{1}{1+\delta} < \dfrac{1}{1-x} < \dfrac{1}{1-\delta}= 2\text{,}$$ thus implying $\left|\dfrac{1}{1-x}\right| < 2$ (notice $\dfrac{1}{1+\delta} = \dfrac{1}{1.5} \in (-2, 2)$, so we can make this claim).

Choose $\delta := \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)$.

Clarinetist
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  • Yes, I'm reading your link and working through the problem. – OGC Sep 13 '15 at 19:10
  • Can we also just pick $\delta = \frac{1}{2}?$ Instead of the minimum. – OGC Sep 13 '15 at 19:21
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    @OGC $\delta$, in some textbooks, is written as $\delta(\epsilon)$, so it has to be a function of $\epsilon$. If I just chose $\delta = 1/2$, I would have $$|x|\left|\dfrac{1}{1-x}\right| < \dfrac{1}{2}(2) = 1$$ but this doesn't give me the desired $\epsilon$ (you aren't allowed to assume $\epsilon > 1$, since it's an arbitrary number $>0$). – Clarinetist Sep 13 '15 at 19:22
  • @OGC I mean, you could probably split it into two cases: $\epsilon > 1$ and $\epsilon \in (0, 1)$, but I'm pretty sure that despite those two cases, you're going to have to make $\delta$ a function of $\epsilon$ somewhere. – Clarinetist Sep 13 '15 at 19:24
  • Sorry I meant $\delta = \frac{\epsilon}{2}$ – OGC Sep 13 '15 at 19:27
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    @OGC I suspect it might work. With $\delta = \dfrac{\epsilon}{2}$, you can show (I leave details to you) that $$\left|\dfrac{1}{1-x}\right| < \dfrac{2}{2+\epsilon} < 2$$ and thus $|x| < \dfrac{\epsilon}{2}$, but I would ask on chat. – Clarinetist Sep 13 '15 at 19:40
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$$\begin{align}\left|\frac{1}{1-x}\right|<\frac{1}{2}\\ \Longleftrightarrow -\frac{1}{2}<\frac{1}{1-x}<\frac{1}{2}\\ \Longleftrightarrow \frac{x-1}{2}<1< \frac{1-x}{2}\\ \Longleftrightarrow x-1<2<1-x\\ \Longleftrightarrow x<3<2-x\end{align}$$ Which is clearly not true for an $x$ arbitrarily near zero. I think there could be a problem with your choice of $\delta$.

nathan.j.mcdougall
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  • Any suggestion for me to pick the right $\delta$? I'm reading Clarinetist's link meanwhile and working through the problem. – OGC Sep 13 '15 at 19:07
  • @OGC $\frac{1}{1-0}=1$, so we'd expect $\left|\frac{1}{1-x}\right|<c$ for some $c>1$. The easiest $c$ to work with is $2$, but you could choose others. Then $-c<\frac{1}{1-x}<c$, $x-1<\frac{1}{c}<1-x$. In the first case $$x-1<\frac{1}{c}\implies x<\frac{1}{c}+1$$ which is clearly true for near-zero $x$. Likewise, $$\frac{1}{c}<1-x\implies x<1-\frac{1}{c}$$ can't be true for fractional $c$ where $x$ is near-zero. If you choose $c=2$, it's obviously true though. So any choice of $c>1$ should suffice. – nathan.j.mcdougall Sep 13 '15 at 19:25