Using delta-epsilon definition of continuous The question is to prove that $f(x) = 2x$ is continuous at $x = 4$ using the delta-epsilon definition of continuous.
So far, I have that we want a $>0$ such that $|x-4|< $ implies that $|2x - 8| < $.
Given $ > 0$ $|2x-8| = 2|x - 4| < 2$.
I am unsure if I have the right approach/where to go from here. Any help would be much appreciated.
 A: Yes, this is the point. You only have now to specify that for the given $\epsilon$ you choose any $\delta$ such that
$$
0<\delta<\epsilon/2
$$
A: We don't know $\delta$ yet. We started with
$$|f(x)-f(4)|=|2x-8|=2|x-4|<\epsilon\implies |x-4|<\frac{\epsilon}{2}.$$
Now, choose $\delta$. Then show that your choice is ok.
A: It helps to recognize the following result from Real Analysis.
Suppose that $S$ is any fixed positive constant (i.e. positive Real number), and $w$ is any variable that is restricted to Real numbers.
Then, you have the result that
$$|w| < S \iff -S < w < S. \tag1 $$
(1) above is an extremely useful result for constructing $\epsilon,\delta$ continuity proofs.
Let $f(x) = 2x.$ 
To prove that $f(x)$ is continuous at $x = 4$, you have to prove two distinct results:

*

*The limit, as $x$ approaches $(4)$, of $f(x)$ exists.

*This limit happens to equal $f(4)$.

Consolidating the definition of a limit existing, with the definition of continuity, you have to prove that
$\forall ~\epsilon > 0, ~\exists ~\delta > 0,~$ such that
$$|x - 4| < \delta \implies |f(x) - f(4)| < \epsilon.\tag2 $$
Remark
This is an alteration of the normal definition of a limit existing.  Such a definition has the premise of
$$\color{red}{0 < }~|x - a| < \delta$$
rather than
$$|x - a| < \delta.$$
This alteration is permitted because

*

*clearly $f(x)$ is well defined at $x = 4$.

*The issue is not whether the limit exists as $x$ approaches $(4)$, but rather whether this limit happens to equal $f(4)$.  
$\color{red}{\text{This is the distinction between}}$ 
$\color{red}{\text{showing that a limit exists and showing}}$ 
$\color{red}{\text{that a function is continuous at a specific value}}$.


The principle established in (1) above can be used to ease the analysis in (2) above.
You want to prove that given any $\epsilon > 0$, a $\delta > 0$ can be found such that the following implication holds:
$$\left\{4 -\delta < x < 4 + \delta\right\} \implies 
\left\{8 - \epsilon < 2x < 8 + \epsilon\right\}. \tag3 $$
The idea is that the assertions in the LHS and RHS of (3) above are equivalent to the following respective assertions:
$$\left\{- \delta < x - 4 < \delta\right\}, ~~~
\left\{- \epsilon < 2x - 8 < \epsilon\right\}. \tag4 $$

Finding the appropriate relationship between $\epsilon$ and $\delta$ so that the implication in (3) above holds, is facilitated by the following consideration (in Real Analysis).
The idea is that if $w$ is a variable, restricted to the Real numbers, and $S,T$ are any Real numbers, then you know that (for example)
$$[ ~S < w < T ~] \iff [ ~2S < 2w < 2T ].$$
Therefore, you can immediately conclude that
$$\left\{4 -\delta < x < 4 + \delta\right\} \iff
\left\{8 -2\delta < 2x < 8 + 2\delta\right\}. \tag 5$$
If you compare the result in (5) with the implication that you are trying to prove in (3), you see immediately that the desired implication will be a consequence of establishing the following relationship between $\epsilon$ and $\delta$:
$$\delta = \frac{\epsilon}{2}.$$
