In the proof of $\lim_{x\rightarrow 1}(2x-1) = 1$, why do we choose $\delta=\epsilon/2$?

Question

First of all, I didn't understand why mathematicians call it a proof?is it really a proof?
What are we trying to prove?
It's kind of like we're putting some numbers and the equality just works.

Consider this limit : $$\lim_{x\rightarrow 1}{2x-1} = 1$$ So I try to do the epsilon delta steps,
$$0<|x-1|<\delta$$

$$|2x-1-1|=|2x-2|= 2|x-1|<\epsilon$$
$$|x-1|< \frac {\epsilon}{2}$$

Now we say $\delta = \frac {\epsilon}{2}$ , but why? I don't understand why we make them equal?

Answer 1

We're basically saying that, provided we make $\delta = \frac{\epsilon}{2}$, then that guarantees $f(x)$ will be within $\epsilon$ units of the limit. We could just as easily say $\delta < \frac{\epsilon}{2}$, but that would be extraneous, because we already have the condition that $|x - 1| < \delta$. We could also just as easily say $\delta = \frac{\epsilon}{3}$, which would certainly also guarantee $f(x)$ is close enough to the limit. It's just that we know setting $\delta = \frac{\epsilon}{2}$ will imply that we can get close enough to the limit. Using $\frac{\epsilon}{2}$ is better, because it is the greatest distance $x$ can be from $1$ while still guaranteeing $f(x)$ will be within $\epsilon$ of the limit.

So, if $\delta = \frac{\epsilon}{2}$, then $$0 < |x - 1| < \frac{\epsilon}{2} \implies |2x -1 -1| < \epsilon$$ which is what we're trying to show.

This is necessary, because it lets us know that we can get $f(x)$ as close to the limit as we want, provided $x$ is within half of the distance between $f(x)$ and the limit. It therefore shows that the limit exists, because we can get as close as we want to it.

Answer 2

Another way of saying $$ \lim_{x\to a}f(x)=b\tag{1} $$ is for all $\epsilon\gt0$, there is a $\delta\gt0$ so that for all $x$ such that $|x-a|\le\delta$, we have $|f(x)-b|\le\epsilon$.

To prove that this latter statement is true, we choose an arbitrary $\epsilon\gt0$, then try to find a $\delta$ (based on our choice of $\epsilon$) that satisfies $$ |x-a|\le\delta\implies|f(x)-b|\le\epsilon\tag{2} $$ In the case of $f(x)=2x-1$, and $a=1$ and $b=1$, for a given $\epsilon\gt0$, $\delta=\epsilon/2$ is a sufficiently small $\delta$ to satisfy $(2)$.