Using the epsilon-delta definition of continuity to prove a linear function is continuous at c

Question

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.

$$h(x)=3x-1$$

Now, for a polynomial, I'd follow three steps:

1. Use algebraic manipulation to express the difference $h(x) − h(c)$
   as a product of the form $(x − c)g(x)$.

2. Obtain an upper bound of the form $|g(x)|<=M$, for $|x − c|<=r$,
   where $r>0$ is chosen so that $[c − r, c + r] \subset A$.

3. Use the fact that $|h(x) − h(c)<= M|x − c|$, 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:

$$h(x)=3x-1, c=4$$
$|h(x)-h(4)| < \epsilon$, for all x with $x-4<\delta$
$$|3x-12|<\epsilon$$
$$3|x-4| < \epsilon$$
$$|x-4|<\frac \epsilon3$$
$$\delta = \frac \epsilon3$$

Therefore, h(x) is continuous at 4.

Where have I gone wrong?

Answer 1

Let $\varepsilon>0$, then taking $\delta=\varepsilon/3$ you have $$|x-4|<\varepsilon/3\implies 3|x-4|<\varepsilon \implies |h(x)-h(4)|<\varepsilon $$ So your $\delta(\varepsilon)$ work. By other hand, your approach is completely correct.