One of my first $\delta$-$\epsilon$ proofs, but I don't feel that I'm heading in the right direction. I want to proof that:

$$ \lim_{x \to \frac{1}{2}} \frac{1}{x} = 2 $$

By the limit definition, For every $\epsilon > 0$, I must find a $\delta$, such that for every $x \in \operatorname{Dom}(f)$:

$$ \left| x - \frac{1}{2}\! \right| < \delta \Rightarrow \left|\! \frac{1}{x} - 2\right| < \epsilon $$

I assume $f(x) = \frac{1}{x}$ is a function $f : \mathbb{R}\backslash \{0\} \rightarrow \mathbb{R}$, since $\frac{1}{x}$ is a reciprocal.

Now, $$ \begin{align} \left|\!\frac{1}{x} - 2\right| &= \left|\!\frac{1}{x} - 2\right|\\ &= \frac{|1-2x|}{|x|} \\ &= \frac{|\!\frac{1}{2}-x|}{2|x|} \end{align} $$

Since $|x-y| = |y-x|$ and therefore,

$$ < \frac{\delta}{2|x|} \\ $$

I recognise that I have to take $\delta<\frac{1}{2}$ to prevent dividing by zero. But finding an estimate in terms of $\delta$ for $\frac{1}{2|x|}$ seems difficult. My attempt:

$$ 2|x| = 2\left(x - \frac{1}{2} + \frac{1}{2}\right) < 2\left|x-\frac{1}{2}\!\right|+1 = 2\delta+1 $$

Concluding, $$ \left|\!\frac{1}{x} - 2\right| < \frac{\delta}{2\delta+1} < \epsilon $$ And therefore, $$ \delta < \min\left(\frac{1}{2}, \frac{1}{\frac{1}{\epsilon} - 2}\right) $$

Is this conceptual proof right? I have the feeling that I'm missing something.

  • 1
    $\begingroup$ I have only read the beginning. You need to go back and edit through. We're talking about $x\in\Bbb R$. No $\Bbb R^n$ when you're talking about the reciprocal. Get rid of all the $d(x,y)$ and $\|x-y\|$, and use plain absolute values everywhere. One further, important hint: You need to bound $x$ away from $0$, so start by assuming $\delta\le 1/4$. $\endgroup$ Jun 20, 2014 at 22:06
  • $\begingroup$ Did that. It seems that bounding away $x$ is helpful. Now, $\frac{1}{2x}$ has a maximum of $2$, therefore $\delta<\frac{\epsilon}{2}$, and $\delta = \min( \frac{1}{4},\frac{\epsilon}{2} )$? $\endgroup$ Jun 20, 2014 at 22:56
  • $\begingroup$ There's still an algebra error. $1-2x=2(\frac12-x)$. But, yes, once you fix that, I believe you should have it right! $\endgroup$ Jun 20, 2014 at 23:13

1 Answer 1


The one equation should be \begin{equation} \frac{|1-2x|}{|x|} = \frac{2|\frac{1}{2}-x|}{|x|}\ . \end{equation} So you have that \begin{equation} d(x,\frac{1}{2}) < \delta \Rightarrow d(\frac{1}{x},2) < \frac{2 \delta}{|x|}\ . \end{equation}

Then, the second problem I see is that you take $2|x| < 2\delta +1$ and conclude that \begin{equation} \frac{\delta}{2|x|} < \frac{\delta}{2\delta +1} \end{equation} The last inequality has to be $>$ since you devide by something larger.

  • $\begingroup$ One more remark. For a rigorous proof I would make a case distinction between $\epsilon \geq \frac{1}{2}$ and $\epsilon < \frac{1}{2}$. $\endgroup$
    – steffens21
    Jun 23, 2014 at 19:50

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