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This question already has an answer here:

Prove using the epsilon-delta definition?

$\lim \limits_{x \to 2}\frac{1}{x} = \frac{1}{2}$

$|f(x)-L|<\epsilon$

$|\frac{1}{x}-\frac{1}{2}|<\epsilon$

$-\epsilon<\frac{1}{x}-\frac{1}{2}<\epsilon$

$-\epsilon+\frac{1}{2}<\frac{1}{x}<\epsilon + \frac{1}{2}$

Any tips on how do I continue from here?

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marked as duplicate by user99914, paf, Gibbs, Shailesh, Ethan Bolker Jun 5 '18 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You need to relate to $|x-2| < \delta$ so I recommend you combine the fractions $\left|\frac{1}{x}-\frac{1}{2}\right|=\left|\frac{2-x}{2x}\right|$ $\endgroup$ – N8tron Jun 4 '18 at 18:14
  • $\begingroup$ @N8tron thanks. I read the proof by John Ma, how do you control the numerator? $\endgroup$ – user567109 Jun 4 '18 at 18:21
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    $\begingroup$ The numerator is controlled by the assumption that $|x-2|< \delta$ the denominator is more interesting. If you for instance assume that $|x-2|<\delta <1$ then $1<x<3$ and $2<2x<6$. $\endgroup$ – N8tron Jun 4 '18 at 18:57
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We can safely assume $0 < \epsilon < \frac 12$ so we can assume $x > 0$ and $-e + \frac 12 > 0$ so

$\frac 1{-\epsilon + \frac 12} > x > \frac 1{\epsilon + \frac 12}$

Now $\frac 1{\epsilon +\frac 12} = \frac 2{2\epsilon + 1}= \frac {4\epsilon +2 -4\epsilon}{2\epsilon + 1}= 2 - \frac {4\epsilon}{2\epsilon +1}$

And $\frac 1{-\epsilon +\frac 12} = \frac 2{-2\epsilon + 1}= \frac {-4\epsilon +2 +4\epsilon}{-2\epsilon + 1}= 2 + \frac {4\epsilon}{-2\epsilon +1}$

So $2 - \frac {4\epsilon}{2\epsilon +1} < x < 2 + \frac {4\epsilon}{-2\epsilon +1}$

Notice $\frac {4\epsilon}{2\epsilon + 1} < \frac {4\epsilon}{-2\epsilon + 1}$ so

$2 - \frac {4\epsilon}{2\epsilon +1} < 2 + \frac {4\epsilon}{2\epsilon +1}< 2 + \frac {4\epsilon}{-2\epsilon +1}$

So IF $2 - \frac {4\epsilon}{2\epsilon +1}< x < 2 + \frac {4\epsilon}{2\epsilon +1}< 2 + \frac {4\epsilon}{-2\epsilon +1}$ THEN $2 - \frac {4\epsilon}{2\epsilon +1} < x < 2 + \frac {4\epsilon}{-2\epsilon +1}$

So If $\delta = \frac {4\epsilon}{2\epsilon +1}$ then

$|x - 2| < \delta$ then

$2 - \frac {4\epsilon}{2\epsilon +1}< x < 2 + \frac {4\epsilon}{2\epsilon +1}$ then

$2 - \frac {4\epsilon}{2\epsilon +1} < x < 2 + \frac {4\epsilon}{-2\epsilon +1}$ then

$|\frac 1x - \frac 12| < \epsilon$.

And we are done.

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I personally find it easier to go forward rather than backwards.

We want $|x - 2| < \delta \implies |\frac 1x - \frac 12| < \epsilon$

$|x - 2| < \delta \implies$

$-\delta < x - 2 < \delta$

$2 - \delta < x < 2 + \delta$ (assuming $x > 0$ and $\delta < 2$)

$\frac 1{2 + \delta} < \frac 1x < \frac 1{2-\delta}$

$\frac 12 - \frac {\delta}{2 + \delta} < \frac 1x < \frac 12 + \frac{\delta}{2-\delta}$.

So now we need

$\frac 12 - \epsilon \le \frac 12 - \frac {\delta}{4 + 2\delta} < \frac 1x < \frac 12 + \frac{\delta}{4-2\delta}\le \frac 12 + \epsilon$.

What $\delta$ will make that work?

Well we need $\epsilon \ge \frac {\delta}{4 + \delta}$ and $\epsilon \ge \frac{\delta}{4-2\delta}$ and so we need $\epsilon \ge \frac{\delta}{4-2\delta} > \frac {\delta}{4 + 2\delta}$

so $\epsilon = \frac{\delta}{4-2\delta}$ will do.

So $4\epsilon - 2\epsilon\delta = \delta$

$4\epsilon = \delta(1 + 2\epsilon)$

$\delta = \frac {4\epsilon}{1 + 2\epsilon}$ is the $\delta$ we need.

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