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

Question

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.

Answer 1

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

Answer 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.

Answer 3

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.