I want to use the $\epsilon-\delta$-criterion to show that for any $c\in\mathbb R$ the function $$ f: \mathbb R \rightarrow \mathbb R, x\mapsto \begin{cases} \frac{1}{x}, &x\neq 0 \\ c, & x=0. \end{cases} $$ is not continuous.
Let $x = 0$. Then we have $$ \mid f(x) - f(0)\mid = \mid\frac{1}{x} - c\mid = \mid \frac{1-xc}{x}\mid = \mid \frac{cx-1}{x} \mid = \frac{\mid cx-1 \mid }{\mid x \mid}. $$ From triangle inequality we have $$ \frac{\mid cx-1 \mid }{\mid x \mid} \leq \frac{\mid cx\mid +1 }{\mid x \mid} = \frac{\mid c\mid \cdot \mid x-0\mid +1 }{\mid x \mid} \leq \frac{\mid c\mid \cdot \delta +1 }{\mid x \mid} $$ But how can I get rid of the denominator? If I can't, what is an argument that it is not possible?
An easier way would to be argue by sequential definition of continuity. If we can find a sequence ${x_n}\to 0$ but $f(x_n)\not\to f(0)=c$ then we're done. But if you pick the sequence $x_n=\frac{1}{n}$ , that converges to zero , but $f(x_n)\rightarrow +\infty$ . Therefore, not continuous at $x=0$, we're done because we found such a sequence.
But if your hand is forced on $\epsilon - \delta$ I think the error in your method comes from trying to estimate $|f(x)-f(0)|$ from above: Really , if you are proving discontinuity, surely you want to try and bound it from below?
Remember , the definition of being discontinuous at $x=0$ is that:
$$\exists \epsilon>0, \forall \delta>0 , \exists x \text{ s.t. } |x|<\delta \text{ but } |f(x)-f(0)|\ge \epsilon$$