Delta-Epsilon proof of $\lim_{x \to a} 2x = 2a$

Question

Prove

$\lim_{x \to a} 2x = 2a$ Using the formal proof, not informal.

So we know

$2|x - a| < \epsilon$

We need to find some $\delta$

We only need to prove there IS SOME $\delta$ right? Only ONE? Is that the definition.

Prove:

$|x - a| < \delta$

We found out.

$2|x - a| < \epsilon \implies |x - a| < \epsilon/2$

So We have,

$|x - a| < \delta$ AND $|x - a| < \epsilon/2$

So $\delta = \epsilon/2 \space \space \space \space \space \space \blacksquare$

This completes the proof.

Is this the way to do it? I have seen though, places where you first Assume $|x - a| < 1$ for example.

That would be correct too right, because all we need it ONE $\delta$ according to the definition.

So,

Assume $|x - a| < 1 \space \space \space \space \space$ AND $|x - a| < \epsilon/2$

But then why is the assumption that $|x - a| < 1$ for example required? Thanks!

Answer 1

This assumption is helpful when you want to find a limit of a more complicated function. I don't think you need that in this case. Generally you can do this because in a limit we are interested in finding a $\delta(\epsilon)$ for a "small" region of $x$. So you can suppose that $|x-a|<1$ or for example you can suppose that $|x-a|<\frac{1}{2}$ and then to determine $\delta$ by setting $\delta=\min\{\delta(\epsilon),1\}$ or $\delta=\min\{\delta(\epsilon),\frac{1}{2}\}$ in the second case. Hence, it not nessecary to assume that but you can do whenever you believe that it can help you. Your solution was right.