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I do not understand what is wrong with this $\epsilon-\delta$ proof that is supposed to show $\displaystyle\lim_{x\to 1}\frac{1}{x} = 1$

We have to show that for every $\epsilon > 0$ there exists a $\delta$ such that for all $x$ that satisfy $0 < \left| x - 1\right| <\delta$ the following is implied: $\left|\dfrac{1}{x} - 1\right| < \epsilon$

Here is how I started: $$ \left|\dfrac{1}{x} - 1\right| < \epsilon \implies\dfrac{\left|x-1\right|}{\left|x\right|} < \epsilon\\ $$ To bound the $\left|x\right|$, let $\delta = 1$ then $$\begin{align*} & \left|x - 1\right| < 1\\ \implies& |x| - |1| \leq \left|x - 1\right| < 1\\ \implies& |x| < 2\\[1em] \implies& |x-1|<\epsilon|x|<2\epsilon \end{align*}$$ So if we let $\delta = \min\left(1,\, 2\epsilon\right)$, all $x$ that satisfy $0<|x - 1|< \delta$ should also satisfy $\left|\dfrac{1}{x} - 1\right| < \epsilon$ for all $\epsilon > 0$, but that is not the case if you take a look at a graph.

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  • $\begingroup$ The last statement is false. Observe that for $\epsilon=1/3$ we have that $0<|x-1|<2/3\implies x\in(-1,1/3)\cup(1,5/3)$ and, at the same time, we have that $\frac{|x-1|}{|x|}<1/3$, but choosing $x=1/4$ we have that $3<1/3$, what is not possible. $\endgroup$ – Masacroso Dec 1 '16 at 7:21
  • $\begingroup$ Your order of conclusion is wrong from the first line (that would be right if you put an \iff $\iff$ there for equivalency). In the end the proof of continuity requires exactly the reverse direction in all of the arrows, and not all of the transformations are to equivalent inequalities. $\endgroup$ – Lutz Lehmann Dec 1 '16 at 12:31
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$\left|\dfrac{1}{x} - 1\right| < \epsilon \implies\dfrac{\left|x-1\right|}{\left|x\right|} < \epsilon$

Great up to here

To bound the $|x|$, let $δ=1$ uh oh...look at what follows

$|x−1|<1\\ 0< x<1\\ \frac {|x-1|}{x} < \infty$

Not going to work. You need a tighter bound for $\delta$

let $\delta = \frac 12$

$|x|> \frac 12\\ |\frac {1-x}{x}| < 2|1-x|<2\delta<\epsilon\\ \delta = \min(\frac 12, \frac \epsilon 2)$

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  • $\begingroup$ I see, if I let $\delta$ equal any value less than 1, it will work? It seems to work for $\dfrac{99}{100}$ $\endgroup$ – E7_82_8E Dec 1 '16 at 7:23
  • $\begingroup$ It will work so long as you can create a lower bound for $x$ that is greater than $0$ $\endgroup$ – Doug M Dec 1 '16 at 7:30
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The inequality $$\vert x - 1 \vert < \epsilon \vert x \vert$$ doesn't follow from the previous line. You need an upper bound on $\vert x \vert$. To get that, just pick $\delta = \min(\epsilon/2,1/2)$. Then $\vert x - 1 \vert < \delta$ implies that $\vert x \vert > 1/2$, so we have $$\frac{\vert x - 1 \vert}{\vert x \vert} < \frac{\epsilon/2}{1/2} = \epsilon.$$

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    $\begingroup$ I can see that letting $\delta = 1$ allows $\frac{1}{x}$ to get arbritarily large, which went against the whole point of bounding $|x - 1|$, but how does $|x - 1| < \epsilon|x|$ not follow from $\dfrac{|x-1|}{|x|}< \epsilon$? Multiplying both sides by $|x|$ results in $|x - 1| < \epsilon|x|$. Is that not legal? $\endgroup$ – E7_82_8E Dec 1 '16 at 7:20

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