Is it possible to disprove continuity using $\epsilon -\delta$ approach?

Question

I am trying to prove that the function $$f(x)=\left\{\begin{array}{ll} 5 x-2 & \text { if } x \neq 1 \\ -2 & \text { if } x=1 \end{array}\right.$$ is Discontinuous at $x=1$ using $\epsilon -\delta$ method.

Since $f(1)=-2$ ,we need to show that, there exists a $\delta >0$ such that $x \in (1-\delta,1)\cup(1,1+\delta)$ $\implies$ $|f(x)+2|>\epsilon$

But not able to proceed?

Answer 1

To show continuity, you have to find a $\delta$ for any $\epsilon$.

To show discontinuity, you have to find an $\epsilon$ for which there is no $\delta$.

The gap between $\lim_{x \to 1} 5x-2 = 3$ and $f(1) = -2$ is $5$. So for any $\epsilon < 5$ it should be impossible to find a $\delta$ to satisfy the requirement.

Answer 2

What you need to show is that, for some $\varepsilon>0$, for every $\delta>0$ there is some number $x$ such that $|x-1|<\delta$ and that $|f(x)-f(1)|\geqslant\varepsilon$.

Take $\varepsilon=4$. Now, for any $\delta>0$, take any $x$ such that $0<|x-1|<\delta$ and also that $|x-1|<\frac15$. Now, note that\begin{align}|x-1|<\frac15&\iff\frac45<x<\frac65\\&\iff2<5x-2<4\\&\iff2<f(x)<4\\&\iff4<f(x)-f(1)<6.\end{align}So, $|f(x)-f(1)|\geqslant1=\varepsilon$.