# Show that limit does not exist ($\varepsilon$-$\delta$)

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 $$$$f(x)= \begin{cases} x,&x>1\\ 3-x,&x\leq1, \end{cases}$$$$ show that $$\lim_{x\to1}f(x)$$ does not exist.

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.

• 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. – bkarpuz Oct 25 '18 at 20:33
• 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 $L$... – bkarpuz Oct 25 '18 at 20:45
• @bkarpuz yes that is a far easier approach! – JP McCarthy Oct 26 '18 at 6:20

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'.

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.

• The question asked for proving that via $\epsilon$-$\delta$. – KM101 Oct 25 '18 at 14:53

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.