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