Is this epsilon-delta proof that limit is not a number correct?

Question

I'd like to proof the following statement wrong by using the epsilon-delta definition:

\begin{equation} \lim_{x\to 2}(x-2) = 1 \end{equation}

\begin{equation} \lim_{x\to 2}(x-2) = 1 \iff \forall \varepsilon > 0 ~ \exists \delta > 0 ~:~ 0 < |x-2| < \delta \implies |x-2 -1|<\varepsilon \end{equation}

I apply the triangle inequality

\begin{equation} |x-2| = |x-2+1-1| = |x-3+1| \leq |x-3| + 1 < \varepsilon+ 1 \end{equation}

And I reach

\begin{equation} |x-2| < \varepsilon+ 1 \end{equation}

Am I in the right way?

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EDIT

I tried a different approach, could you check if it is now correct??

I'm assuming that the following statement is true and try to reach a contradiction.

\begin{equation} \lim_{x\to 2}(x-2) = 1 \iff \forall \varepsilon > 0 ~ \exists \delta > 0 ~:~ 0 < |x-2| < \delta \implies |x-2 -1|<\varepsilon \end{equation}

I assume $0 < |x-2| < \delta $. Let $\varepsilon = 0.5$. Then by assumption it is true:

$|x-2-1| <\varepsilon= 0.5 $

However, I can't find any $\delta$ such that $0<|x-2|<\delta$ implies $|x-2-1| < 0.5$. This is because if $x$ lies in $(2-\delta,2+\delta)$, then this would have to impliy that $x$ lies in $(2.5,3.5)$ which is not true.

Now I'm asking, is my reasoning correct? If not why not?

(BTW I had already tried other ways to prove this, just wanted to try a different approach)

Answer 1

Suppose I hypothesize $$\lim_{x\to a} f(x) = L$$ To show this is false, I want to prove the negation of it's definition. The definition is: $\forall \epsilon > 0 \exists \delta > 0 \forall x: \;(0 < |x - a| < \delta) \implies |f(x) - L| < \epsilon$, so it's negation would be $\exists \epsilon > 0\forall\delta>0\exists x:( 0 < |x - a| < \delta) \land|f(x) - L| \geq \epsilon$

For your example, choose $\epsilon = \frac{1}{2}$. Now for any $\delta > 0$, we know that $|x-2| < \delta \implies -|x-2| > -\delta$.

Choose $x=2+\min(\frac{\delta}{2},\frac{1}{2})$ so that $|x-2|< \delta$. At this $x$, if $\delta \geq 1$, we have: $$|x-3| = \frac{1}{2}\geq\epsilon$$ If instead $\delta < 1$, then $-\frac{\delta}{2}>-\frac{1}{2}$ we have $$|x-3| = \left|\frac{\delta}{2}-1\right| \geq 1 - \frac{\delta}{2} > 1 - \frac{1}{2} = \frac{1}{2} = \epsilon$$ And so we've proven the negation i.e $$\exists \epsilon=\frac{1}{2}\forall \delta>0 \exists x=2+\min\left(\frac{\delta}{2},\frac{1}{2}\right): |x-2|<\delta \land |x-2-1| \geq \epsilon \tag*{$\blacksquare$}$$

Answer 2

To show the limit is not equal to $1$, it suffices to prove the negation... that is, that there exists some $\epsilon>0$ such that for every $\delta>0$ we can always find some $x$ with $|x-2|<\delta$ such that $|(x-2)-1|>\epsilon$

Let us take $\epsilon = \frac{1}{2}$

Then, whatever $\delta$ happened to be, let us take $x = \min(2.1,2+\frac{\delta}{2})$. We use this minimum here as a just in case $2.1$ fell outside the range.

Then it is clear that $|x-2|<\delta$ as well as that $|(x-2)-1|\geq 0.9>0.5=\epsilon$