# Proving the limit $\lim_{x \to 2}\frac{1}{x}=\frac{1}{2}$

I wish to prove the limit $$\lim_{x \to 2}\frac{1}{x}=\frac{1}{2}$$

In other words, given $\epsilon > 0$, I wish to prove that I can find a $\delta > 0$ so that $$|x - 2| < \delta \implies |\frac{1}{x}-\frac{1}{2}|< \epsilon$$

Start out by noting that by taking the multiplicative inverse of everything on the right side of the implication sign, we have that $|x-2|<\frac{1}{\epsilon}$. Choose $\delta = \min\{1, \frac{1}{\epsilon}\}$. We have that $$|x-2|<\frac{1}{\epsilon}\implies|\frac{1}{x}-\frac{1}{2}|<\frac{1}{\frac{1}{\epsilon}} = \epsilon$$

And I claim we are done. The answer differs slightly from the one in the text, so I would like some confirmation/slaughter.

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"take the multiplicative inverse of everything"?? –  user2345215 Mar 6 '14 at 10:02

False. $$|x-2|<\frac{1}{\epsilon}$$ is not the multiplicative inverse of $$\left|\frac1x - \frac12\right|<\epsilon.$$ The inverse of $\left|\frac1x - \frac12\right|$would be $$\left|\frac{1}{\frac1x-\frac12}\right|=\left|\frac{2x}{2-x}\right|,$$ not what you got.

However, it does not even matter what the inverse is. If you have $0<a<b$, what you can conclude is $\frac1a > \frac 1b$, so even if your inverse would be correct, you would end up with an inequality facing the other way.

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Oh, jeez, that's not even funny. –  Andrew Thompson Mar 6 '14 at 10:04
A quick hint why I (and most other math students past their first year) noticed things went wrong: The $\delta$ you find always becomes smaller when $\epsilon$ becomes small. Your selection $\delta = \epsilon^{-1}$ meant that as $\epsilon$ was small, you could just take any $\delta$ you wanted, as long as it was smaller than $1$. This is clearly wrong. –  5xum Mar 6 '14 at 10:08

If I understand your solution correctly, there is a problem with taking "the multiplicative inverse of everything". Consider, for example, $|2+3|<6$, which does not imply $|1/2 + 1/3| < 1/6$.

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You can't start by manipulating the thesis. You must start from $|x-2|<\delta$. So: $$|x-2|<\delta\Rightarrow |2x||\frac{1}{x}-\frac{1}{2}|<\delta\Rightarrow |\frac{1}{x}-\frac{1}{2}|<\frac{\delta}{2|x|},$$ but $x>\delta-2$, so: $$|\frac{1}{x}-\frac{1}{2}|<\frac{\delta}{2|x|}<\frac{\delta}{2(\delta-2)},$$ write then $$\varepsilon=\frac{\delta}{2(\delta-2)}.$$

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Sorry, I had to use the other inequality, I'm correcting it right now. –  7raiden7 Mar 6 '14 at 10:12

While I did accept the answer giving the best response, I feel like I have to solve it correctly after posting such an embarassing blunder. We wish to prove the limit $$\lim_{x \to 2} \frac{1}{x} = \frac{1}{2}$$ In other words, we wish to show that for every $\epsilon > 0,$ we can find a $\delta > 0$ so that $$|x - 2| < \delta \implies |\frac{1}{x} - \frac{1}{2}| < \epsilon$$

Manipulating the expression $|\frac{1}{x}-\frac{1}{2}| = |(2-x)\frac{1}{2x}| = |(x-2)\frac{1}{2x}|$. Let $\delta < 1$. We have that $-1 < x - 2 < 1 \implies 1 < x < 3$, and, learning from past mistakes, $|\frac{1}{2x}|<\frac{1}{2}$. (On the exam in December where I initially got this task, I ended up writing $|\frac{1}{2x}| < \frac{1}{6}$, so you can see my misunderstanding with "multiplicative inverses" is something that has been with me for a while. I have never even thought it through, even though I see it as, ignorant as it might sound, obvious now that someone actually pointed it out.) Choosing $\delta = \min\{1, 2\epsilon\}$, we have that $$|x - 2| < \delta \implies |\frac{1}{x} - \frac{1}{2}| < \frac{1}{2}\cdot2\epsilon = \epsilon$$

And we are done.

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