Proving Discontinuity using Epsilon Delta for Function

Question

sorry if this question ends up being a duplicate, but I wasn't able to find a question sufficiently similar on the exchange.

I have a few questions to go through to practice proving functions are discontinuous using the delta epsilon definition of a limit. I am familiar with using delta-epsilon to prove functions are continuous at a point, but not that they are discontinuous in general. In order to help me solve later problems, please provide a full solution to the problem below, explained as clearly as possible. Thank you! The problem is:

Show that the following function is discontinuous for all $c \in \mathbb R$:

$f(x) = \begin{cases} 1/x, & \text{$x \ne 0$} \\ c, & \text{$x = 0$} \end{cases}$

Answer 1

Here is a hint which you can play around with to aid your understanding:

Let $\delta > 0$. Notice that if you choose $x$ such that $0 < |x| < \min\left(\delta, \frac{1}{|c|+1}\right)$ then

$$ |f(x) - c| = \left|\frac{1}{x} - c \right| = \left|\frac{1-cx}{x} \right| \geq \frac{1 - \frac{|c|}{|c|+1}}{\frac{1}{|c|+1}} = 1 $$

Answer 2

Since $1/x$ is continuous everywhere it's defined, we expect $f$ will be discontinuous at $0$. SO, suppose on contrary that $f$ is continuous at $0$. Then, we know that for all $\epsilon>0$ there is a $\delta>0$ such that $|x|<\delta\implies|c-f(x)|<\epsilon$. So, if we choose $\epsilon=1$, then there is a $\delta>0$ such that $|x|<\delta\implies|c-f(x)|<1$. Now since $1/x$ blows up at $0$, we can choose an $0<x_{0}<\delta$ such that $1/x_{0}>|c|+1$. Then, $|x_{0}|<\delta$, but $|c-f(x)|=|c-1/x_{0}|\geq 1/x_{0}-|c|>1\not<\epsilon$. This contradiction proves the desired discontinuity.

Answer 3

We want to prove the discontinuity at $0$, i.e. $$\exists\epsilon>0,\forall\delta>0,\exists x\in[-\delta,\delta]\setminus\{0\},\left|\frac1x-c\right|\ge\epsilon.$$ For this, any $\epsilon>0$ will do, as it is then sufficient, for every $\delta>0$, to find some $x\in(0,\delta]$ such that $\frac1x\ge|c|+\epsilon$, i.e. such that $x\le\frac1{|c|+\epsilon}$, and there are many such $x$s: the whole interval $\left(0,\min\left(\delta,\frac1{|c|+\epsilon}\right)\right]$.