It has been a while since I have taken an analysis course and I have a question regarding continuity at a point. I am working on the following question.
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined by $$ f(x) = \left\{ \begin{array}{ll} x+1 & \quad x \geq 0 \\ x-1 & \quad x < 0 \end{array} \right. $$ Prove that $f$ is continuous at $c$ when $c \neq 0$.
The definition of continuity at a point $c$ is defined as:
$\forall \epsilon >0 \exists \delta >0 \forall x \in \mathbb{R} (|x-c|<\delta \Rightarrow |f(x)-f(c)|<\epsilon)$.
proof attempt:
Consider two cases: (1) $c<0$, (2) $c>0$.
(1) If $c<0$, then we are looking at all $x<0$. $|f(x)-f(c)|=|x-1-f(c)|$ and we want this to be less than $\epsilon$. However, this is where I am getting stuck and I am not sure how to continue. I feel as though case (2) will be very similar.
Any insight/hints in the right direction are appreciated, please no full solution though, I would like to understand this fully myself. Thank you.
Hint:
For case 1 ($c<0$):
Let $\varepsilon > 0$.
Choose $\delta = \min \left\{ { - \frac{c}{2},\frac{\varepsilon }{2}} \right\}$
Let there be $x \in \Re$ such that $\left| {x - c} \right| < \delta $.
Notice that $$\left| {x - c} \right| < - \frac{c}{2}$$ hence $$\frac{c}{2} < x - c < - \frac{c}{2}$$ therefore $$x < \frac{c}{2} < 0$$ Now look at $$\left| {f\left( x \right) - f\left( c \right)} \right| = ...$$
Can you continue from here?