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

continuityepsilon-delta

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?

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.