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I'm trying to prove the following using the epsilon-delta method:

$$\lim_{x\to 1}\frac{1}{x} = 1$$

Prior to reading the fantastic answer by @Clarinetist in this question, I tried writing the proof myself. My approach begins similarly to the aforementioned response, but I wanted to verify the final details of it. Anyway, here it is:

$$\left|\frac{1}{x} - 1\right| = \left|\frac{1-x}{x}\right| = \left|\frac{x-1}{x}\right|$$ $$ = \frac{|x - 1|}{|x|}$$

Now, if $|x - 1| < |x|\epsilon$, we'll be done.

$$|x - 1| < |x|\epsilon\implies |x| - |1| < |x|\epsilon$$ $$|x| - |x|\epsilon < 1\implies |x|(1-\epsilon) < 1$$ $$|x| < \frac{1}{1-\epsilon}$$

Now:

$$|x - 1| < \frac{1}{1-\epsilon} - 1\implies |x| - |1| < \frac{1}{1-\epsilon} - 1\implies |x| < \frac{1}{1-\epsilon}$$

Therefore, $\delta = \frac{1}{1-\epsilon} - 1$

Aryaan
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  • You have implications in the wrong direction. $|x| <\frac 1 {1-\epsilon}$ does not yield $|x-1|<|x| \epsilon$. – Kavi Rama Murthy Jun 26 '24 at 08:03
  • @geetha290krm why not? I would assume that each step is reversible. That is, $|x| < \frac{1}{1-\epsilon}\implies |x|(1 - \epsilon) < 1\implies |x| - |x|\epsilon < 1$ and so on. Could you point out which specific step fails to obey this assumption? – Aryaan Jun 26 '24 at 08:06
  • $|x|-|x|\epsilon <1$ does not imply that $|x-1|<|x|\epsilon$. [Take, for example, $x=0$]. – Kavi Rama Murthy Jun 26 '24 at 08:14
  • @geetha290krm ah my bad. Thanks for the clarification – Aryaan Jun 26 '24 at 08:17

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