# Proof that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon$-$\delta$ definition of limits (verification).

Prove that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon-\delta$ definition of limits.

\\ \begin{align} \\ &\textrm{Let } \forall \epsilon > 0 \\ &\textrm{Choose } \delta = \min{\{1, 2\epsilon \}} \\ &\textrm{Assume } 0 < |x - 2| < \delta : \\ \end{align}

\\ \begin{align} \\ \left|\frac{1}{x} - \frac{1}{2}\right| &< \epsilon \\ \frac{|2 - x|}{|2x|} &< \epsilon \\ |-1(x - 2)| &< \epsilon|2x| \\ |x - 2| &< \epsilon|2x| \\ \end{align}

\\ \begin{align} \\ |x - 2| &< 1 \\ -1 < x - 2 &< 1 \\ 1 < x &< 3 \\ \end{align}

\\ \begin{align} \\ |x - 2| &< \epsilon|2(1)| \\ |x - 2| &< 2\epsilon \\ \end{align}

$$\\\therefore \delta \leq 2\epsilon$$

Right, so I start by taking $|f(x)−L|<ϵ$. I then isolate $|x−2|$ to the left. I then limit $|x−2|$ to be less than one and then find a range of $x$ values which satisfy the inequality. Then I plug in the smallest $x$ value to minimise the value of $ϵ$. I make the conclusion that $δ≤2ϵ$. Am I excluding or misplacing steps? I'm fairly new to this whole thing.

• Use some words, I can't understand what you're doing. Sometimes it seems you're writing down equivalences, then you jump to a side argument hoping to ascertain some fact (I presume), then you somehow get back into the proof. All in all, I don't understand it. Feb 14, 2015 at 12:02
• @GitGud Right, so I start by taking $|f(x) - L| < \epsilon$. I then isolate $|x-2|$ to the left. I then limit $|x - 2|$ to be less than one and then find a range of $x$ values which satisfy the inequality. Then I plug in the smallest $x$ value to minimise the value of $\epsilon$. I make the conclusion that $\delta \leq 2\epsilon$. Am I excluding or misplacing steps? I'm fairly new to this whole thing.
– user146046
Feb 14, 2015 at 12:10
• @GitGud Does what I said make for any better clarity? Apologies for the poor structure.
– user146046
Feb 14, 2015 at 12:33
• It helps. Your ideas are correct, the way you wrote the proof is not. For this reason I added the tag (proof-writing). A user posted a very clean proof. This what you should thrive for. You posted scratch work. In this sort of problem, it's very common to just explain the reasoning behind it as you did in your question together with your comment. For many people this is enough. Other people will want, in addition to what you did, that after the scratch work you post a 'clean' solution. Other people only require the 'clean' solution. The latter is my personal choice. Feb 14, 2015 at 12:53
• A very misleading thing about what you typed is that you started out as if you were going to write a clean proof without scratch work, this is due to "$\text{Choose } \delta = \min\{1, 2\epsilon \}$", but then you proceed as if you didn't know what $\delta$ is, but you did and you didn't use this information. Remove the quoted sentence, add your comment to the question and after $\delta \color{red}{<}2\epsilon$ finalize with "so it's enough to take $\delta<2\epsilon$" and what you did becomes "the reasoning behind it" that I mentioned in the prior comment. Feb 14, 2015 at 12:58

Here you have to show that for each $\epsilon >0$, there exists $\delta >0$ such that for each $x\in Domn(\frac{1}{x})$ if $0<|x-2|<\delta$ then $\left | \frac{1}{x}-\frac{1}{2}\right |<\epsilon$.

So begin with arbitrary $\epsilon >0$.

Notice that if $0<|x-2|<1$ then $1<|x|<3$ and hence $\frac{1}{3}<\frac{1}{|x|}<1$.

Now choose $\delta =\min\{1,2\epsilon\}$. Then clearly $\delta >0$.

Now suppose $0<|x-2|<\delta$.

Then $\left | \frac{1}{x}-\frac{1}{2}\right |=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<\frac{2\epsilon}{2}=\epsilon$.

Therefore $$\lim_{x\rightarrow 2}\frac{1}{x}=\frac{1}{2} .$$

The inequalities you have written can be rearranged and filled by implications and quantification to lead to a correct and clear proof. Mathematics is not about writing formulas, but developing a reasoning... at least at higher levels. I suggest you to write the missing part if you want people to understand easily your thought.

Here the problem is to use the $\epsilon-\delta$ machinery in a proper way that we can always use in any situation.

The start point is to give $\epsilon>0$ and to find a solution of

$$\left| \dfrac{1}{x}-\dfrac{1}{2}\right| <\epsilon$$ If this solution can be expressed in the form $|x-2|<\delta$ (with $\delta$ depending in general from $\epsilon$) we have found a $\delta$-neigborought of $2$ that is a solution of the inequality and we have proved the limit.

In our case, since we are interested only to solutions in a neigborought of $2$,solving for the absolute value the inequality become:

$$\begin{cases} 0<x<2\\ 2-x<2\epsilon x \end {cases} \land \begin{cases} x>2\\ x-2<2\epsilon x \end {cases}$$

Solving the first system we find: $$x>\dfrac{1}{1+2\epsilon}$$ that we can put in the form $$x>2-\dfrac{4\epsilon}{1+2\epsilon}$$ i.e.

$2>x>2-\delta_1$ with $\delta_1= \dfrac{4\epsilon}{1+2\epsilon}$

Now we solve the second system and, in the same way, we find:

$$2<x< 2+\dfrac{4\epsilon}{1-2\epsilon}=2+\delta_2$$

and finally, chosing $\delta=$min$(\delta_1,\delta_2)$ we find the $\delta-$neigborought $|x-2|<\delta$ that we was searching.