# Give a delta-epsilon proof that $\lim_{x \to \frac{1}{2}} \frac{1}{x} = 2$, is my answer right?

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.

• 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 $\delta\le 1/4$. Jun 20, 2014 at 22:06
• 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 $\delta = \min( \frac{1}{4},\frac{\epsilon}{2} )$? Jun 20, 2014 at 22:56
• There's still an algebra error. $1-2x=2(\frac12-x)$. But, yes, once you fix that, I believe you should have it right! Jun 20, 2014 at 23:13

The one equation should be $$\frac{|1-2x|}{|x|} = \frac{2|\frac{1}{2}-x|}{|x|}\ .$$ So you have that $$d(x,\frac{1}{2}) < \delta \Rightarrow d(\frac{1}{x},2) < \frac{2 \delta}{|x|}\ .$$
Then, the second problem I see is that you take $2|x| < 2\delta +1$ and conclude that $$\frac{\delta}{2|x|} < \frac{\delta}{2\delta +1}$$ The last inequality has to be $>$ since you devide by something larger.
• One more remark. For a rigorous proof I would make a case distinction between $\epsilon \geq \frac{1}{2}$ and $\epsilon < \frac{1}{2}$. Jun 23, 2014 at 19:50