Epsilon delta proof - why and how we define epsilon in terms of delta I don't understand how the epsilon and delta are determined/decided upon.
For the sake of this question, I need to prove $$\lim_{x\to3}\frac{x-1}{2}=1.$$
After we are done with the algebra, we define $\delta$ as $2\epsilon$. I don't understand how and why we do this.
Thank you, and have a nice day.
 A: The definition of a limit starts with:

For all $\varepsilon > 0$ there exists some $\delta$ such that...

Proving
$$\lim_{x\rightarrow 3}\frac{x-1}2 = 1$$
is sort of like playing a game: first, your opponent chooses some $\varepsilon$ to challenge you with. You win if you can respond with some $\delta$ with the desired property. A proof is just some method to show that, no matter what your opponent says, you can respond with an appropriate $\delta$. A counterexample (or a flaw in your proof) is some $\varepsilon$ for which your opponent would win.
The order of events here is important: when you need to come up with $\delta$, you already know which $\varepsilon$ was chosen. You can think of something like $\delta = 2\varepsilon$ as naming a strategy: no matter what your opponent says, you'll respond with twice $\varepsilon$ and win. Of course, there are other strategies that work (for instance, you could just respond with $\varepsilon$ itself - it also has the desired property) - but the point is that deriving a formula for any appropriate $\delta$ in terms of $\varepsilon$ is always allowable, because $\varepsilon$ is always known before $\delta$ is required.
In this specific case, the algebra will tell you that if $|x-3| < 2\varepsilon$ then $\left|\frac{x-1}2 - 1\right| < \varepsilon$, where $2\varepsilon$ has exactly the property needed - so saying $\delta = 2\varepsilon$ is indeed a winning strategy.

Worthy of note, because it's a common mistake: note that you may never influence what $\varepsilon$ was. It is handed to you by someone who wants to make your proof fail. If you derive a formula for $\varepsilon$ in terms of $\delta$, you have not succeeded. If your proof only works for certain $\varepsilon$, you have not succeeded. Your only move is to choose $\delta$, but you have the advantage of already knowing $\varepsilon$ when you do it.
A: when you write $$ \lim_{x\to a} f(x) = L$$ where L is a finite number. It means that you can always choose a interval arround x=a such that distance between f(x) and L can be minimized as you wish (however small). In your case if you want that distance between $\frac{x-1}{2}$ and $1$ should be less than $\epsilon$ then you can do it when you will select inverval $(3-\delta, 3+\delta)$ where $\delta = 2\epsilon$
A: For a fixed $\varepsilon>0$, we have:
$$\left|\frac{x-1}{2}-1\right|=\frac{1}{2}\left|x-3\right|<\varepsilon\iff|x-3|<2\varepsilon$$
Since we are computing the limit as $x\to3$, if we choose $\delta=2\varepsilon$, then when $|x-3|<\delta$, we also have $|x-3|<2\varepsilon$ by the above computation, and working backwards, we find that when $|x-3|<\delta$, $|(x-1)/2-1|<\varepsilon$, which is what we want in order to show that the limit as $x\to 3$ is $1$.
A: One doesn't "define" $\delta>0$.  
Rather, all one needs is to find any number $\delta>0$ such that for a given $\varepsilon$, $\left|\frac{X-1}{2}-1\right|<\varepsilon$, whenever $0<|x-3|<\delta$.  That is to say that $\delta$ is not unique.  In fact, if we find a number $\delta$, then any number less than that $\delta$ would also qualify.
In the case of interest, the value $\delta =2\varepsilon$ is actually the largest value of $\delta$ that works.  So, in this case we have the largest interval $x\in(3-2\varepsilon ,3+2\varepsilon)$ for which $1-\varepsilon<\frac{x-1}2<1+\varepsilon$.  Any number smaller than $2\varepsilon$ would lead to a smaller interval over which we bound $\left|\frac{x-1}2-1\right|$ by $\varepsilon$.
