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By means of $\varepsilon$-$\delta$, I am looking for some ideas to prove (a beginner math class) that limit does not exist. For instance, consider the function \begin{equation} f(x)= \begin{cases} x,&x>1\\ 3-x,&x\leq1, \end{cases} \end{equation} show that $\lim_{x\to1}f(x)$ does not exist.

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I used to find the negation of $\varepsilon-\delta$ very tricky as an undergrad. It appears I still do because I am not 100% sure of this.

I reckon the limit does not exist if for all $\delta>0$ and $L$, there exists an $\varepsilon>0$ and an $x$ such that $0<|x-1|<\delta$ but $|f(x)-L|\geq \varepsilon$.

In this is the correct negation, then for any $\delta>0$, and any $L$, choose $\varepsilon=1/3$. We might have to pick $x$ according to what $L$ is.

Say $$x(L)=\begin{cases} 1+\min\{\delta/2,1/6\}, & \text{ if }L\leq 3/2 \\ 1-\min\{\delta/2,1/6\}, &\text{ if }L>3/2 \end{cases}.$$

Suppose that $L\leq 3/2$. With $x=1+\min\{\delta/2,1/6\}$, we have have $|x-1|=\min\{\delta/2,1/6\}<\delta$, so $x$ is $\delta$-close to one. Then $|f(x)-L|=|L-(1+\min\{\delta/2,1/6\})|\geq 1/3$.

Similarly for $L>3/2$, there exists a point $\delta$-close to $1$, $x=1-\min\{\delta/2,1/6\}$, such that $|f(x)-L|\geq 1/3$.

I wouldn't be one bit surprised however if I have messed up the negation.

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    $\begingroup$ I did similar proof in the class. I used $\varepsilon=1/2$ and showed that there is no $\delta>0$ for $L=1$ and $L=2$. And, mentioned that the proofs for the cases $L<1$, $1<L<2$ and $L>2$ follow similarly. $\endgroup$ – bkarpuz Oct 25 '18 at 20:33
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    $\begingroup$ If for every $\varepsilon>0$ there exists $\delta>0$ such that $0<|x-1|<\delta$ implies $|f(x)-L|<\varepsilon$. Then, this is also true for $\varepsilon=1/2$. And obtain contradiction for every possible $L$... $\endgroup$ – bkarpuz Oct 25 '18 at 20:45
  • $\begingroup$ @bkarpuz yes that is a far easier approach! $\endgroup$ – JP McCarthy Oct 26 '18 at 6:20
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Seems pretty tough for a beginner maths class lol. If you end up with an answer which doesn't satisfy $|x-x_0|<\delta \implies |f(x)-L|<\varepsilon$, then you just say 'limit doesn't exist'.

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It doesn't exist because the left limit at $x = 1$ is $2$ and right limit at $x = 1$ is $1$. If a proper limit existed there, left and right limits would be equal.

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    $\begingroup$ The question asked for proving that via $\epsilon$-$\delta$. $\endgroup$ – KM101 Oct 25 '18 at 14:53
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I guess there's no easy way of doing this. Here's the way I would proceed:

Assume the limit exists and is some number $L$. Then for $\varepsilon =1$, there is an $\delta > 0$ such that $|f(x) - L|<1$ for all $0<|x-1|<\delta$.

Now if $\delta <1$, let's define $\delta ' := \delta$, and otherwise let $\delta '$ be some number with $\delta ' <1 \le \delta$.

Now pick $x_1$ and $x_2$ such that $1-\delta ' < x_1 < 1 < x_2 < 1+ \delta '$.Then it is clear that $0<|x_1-1|<\delta$ and $0<|x_2-1|<\delta$.

Thus, $|f(x_1) -1| < 1$. This means $|3-(x_1 + L)| <1$. Now, by using the fact that $x_1>0$, we can conclude that $|L|>2$. (Check this!)

Also, we have that $|f(x_2) -1| < 1$. This means $|x_2-L|<1$. Then $|L| < 1+ |x_2| < 1+\delta ' <2$.

But we showed $|L|<2$ and $|L|>2$, which is a contradiction. Thus, limit does not exist.

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