I'm familiar with using the definition to prove a polynomial is continuous at point c, but I've yet to use it with respect to linear functions.
Example:
Prove $h(x)$ is continuous at 4.
$$h(x)=3x1$$
Now, for a polynomial, I'd follow three steps:

Use algebraic manipulation to express the difference $h(x) − h(c)$
as a product of the form $(x − c)g(x)$. 
Obtain an upper bound of the form $g(x)<=M$, for $x − c<=r$,
where $r>0$ is chosen so that $[c − r, c + r] \subset A$. 
Use the fact that $h(x) − h(c)<= Mx − c$, for $x − c<= r$, to
choose $\delta > 0$ such that $h(x) − h(c) < \epsilon$, for all $x
\in A$ with $x − c < \delta$.
Now, seeing as it's a linear function, step 2 is irrelevant, and should look something like this:
$$h(x)=3x1, c=4$$
$h(x)h(4) < \epsilon$, for all x with $x4<\delta$
$$3x12<\epsilon$$
$$3x4 < \epsilon$$
$$x4<\frac \epsilon3$$
$$\delta = \frac \epsilon3$$
Therefore, h(x) is continuous at 4.
Where have I gone wrong?
Best Answer
Let $\varepsilon>0$, then taking $\delta=\varepsilon/3$ you have $$x4<\varepsilon/3\implies 3x4<\varepsilon \implies h(x)h(4)<\varepsilon $$ So your $\delta(\varepsilon)$ work. By other hand, your approach is completely correct.