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

limits

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.

Best Answer

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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