Linear Function is Continuous/Proof 1

Proof
First assume $\alpha \ne 0$.

Let $\epsilon > 0$.

Let $\delta = \dfrac \epsilon {\size \alpha}$.

Then, provided that $\size {x - c} < \delta$:

So, we have found a $\delta$ for a given $\epsilon$ so as to make $\size {\map f x - \map f c} < \epsilon$ provided $\size {x - c} < \delta$.

So $\displaystyle \lim_{x \to c} \map f x = \map f c$ and so $f$ is continuous at $c$, whatever $c$ happens to be.

Now suppose $\alpha = 0$.

Then $\forall x \in \R: \map f x - \map f c = 0$.

So whatever $\epsilon > 0$ we care to choose, $\size {\map f x - \map f c} < \epsilon$, and whatever $\delta$ may happen to be is irrelevant.

Continuity follows for all $c \in \R$, as above.