# Proving discontinuity by epsilon-delta

Given $$f(x)=\begin{cases} 2x+3, x\ge 1 \\ 2x-2, x\lt 1 \end{cases}$$

Show it is discontinuous at x=1, using epsilon delta. Any pushes in the right direction would be greatly appreciated.

To show explicitly, you must show that there exists an $\epsilon>0$ such that for all $\delta>0$, you can find some $x \in B(1,\delta)$ (that is, $|x-1| < \delta$) such that $f(x) \notin B(f(1),\epsilon)$ (that is, $|f(x)-5| \ge \epsilon$).

If you plot $f$ you will see that $f(x) \le 0$ for all $x < 1$, so we can choose $\epsilon = 2$. Let $\delta>0$ be arbitrary, then $f(1-\frac{1}{2}\delta) \le 0$, and so $|f(1-\frac{1}{2}\delta)-5| \ge 3 > \epsilon$, as required.

Hence $f$ is discontinuous at $x=1$.

We have $f(1)=5$. So to show that $f$ is not continuous at $x=1$, it is enough to show that it is not true that $\lim_{x\to 1} f(x)= 5$.

Suppose to the contrary that the limit exists and is equal to $5$. Then for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-1|\lt\delta$, then $|f(x)-5|\lt\epsilon$.

Pick $\epsilon=\frac{1}{2}$. We show there is no $\delta$ with the required property.

If $x\lt 1$, then $f(x)\lt 1$. In particular, if $x\lt 1$, then $|f(x)-5|\gt 4$. It follows that there is no $\delta$ such that $|x-1|\lt \delta$ guarantees that $|f(x)-5|\lt \epsilon$.

Remark: The idea is basically geometric. We are showing that there are points arbitrarily near $x=1$ at which the function value is not close to $5$. Once one has the geometry under control, writing out the details in $\epsilon$-$\delta$ language is a translation job.

We need to show that there exists a fixed $\varepsilon>0$ such that for any chosen $\delta>0$ there is an $x$ satisfying $|x-1|<\delta$, but $|f(x)-f(1)|\ge\varepsilon$. Then $f$ cannot be continuous at $1$.

Try fixing $\varepsilon=1$. Now let us look for a suitable $x$ candidate. If we choose $x$ such that $x<1$ and $|x-1|<\delta$ then can you see that $f(x)<0$? This inequality follows: $$|f(x)-f(1)|\ge |0-f(1)|=|f(1)|=|5|=5>1=\varepsilon$$ and we are done.

In the case where $$f(1)$$ is not defined I think it is good to follow the method below.

Suppose $$f(x) \in C\{1\}$$ and $$Let f(1)=L$$, then for each $$\varepsilon>0$$ we can find $$\delta>0$$ such that $$0 \le |x-1 | < \delta \Rightarrow |f(x)-L|<\varepsilon \iff - \varepsilon+f(x)\ Let $$f(1- \delta)=2(1-\delta)-2=-2\delta$$. We have $$-\varepsilon-2\delta Similarly, $$f(1+\delta)=2(1+\delta)+3=5+2 \delta$$. We have
$$-\varepsilon+5+2\delta Choosing $$\delta=\varepsilon=1$$ we have $$L \in (-3,-1)$$ and $$L\in(6,8)$$ which is a contradiction.