Circular Reasoning for Epsilon-Delta Proof? $$\lim_\limits{x \to 4} 2x-5=3$$
In order to prove this limit, the epsilon-delta definition will be used.  
$$|f(x)-L|<\varepsilon$$ $$|x-a|<\delta$$
In the proof, the above $2$ inequalities will be used to find how $\delta$ is related to $\varepsilon$ (e.g. $\delta=\epsilon/2$).
Then, this relationship between $\delta$ and $\varepsilon$ will be used show that for any $\varepsilon > 0$,
$$|x-a|<\delta$$
will result in
$$|f(x)-L|<\varepsilon$$
Which seems weird to me, because it doesn't seem to be a good proof as there seems to be circular reasoning, but I am probably missing something here.  
Am I missing out any details which prevents circular reasoning in this proof?
 A: Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.
The point of the proof is this: show that for any number $\epsilon>0$ we could choose, there exists a corresponding $\delta>0$ so that for every $x$ with $|x-a|<\delta$ we have $|f(x) - L|<\epsilon$.
To do this, we choose any arbitrary $\epsilon$ and find a $\delta$ which corresponds to that value of $\epsilon$ to make is so that when we are within $\delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $\epsilon$ of L.
The point is that we pick an arbitrary $\epsilon$ and find a $\delta$ which corresponds to it, thus demonstrating that no matter what $\epsilon$ we pick, we can always find a $\delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other 
A: There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $\delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $\epsilon$/$\delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $\delta$ (in terms of $\epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.
A: I like to teach it this way: Given $\epsilon > 0$, your job is to prove the existence of an appropriate $\delta > 0$. One way you can do that is to use mathematical explorations to figure out an actual formula $\delta = \delta(\epsilon)$ expressed as a function of $\epsilon$. You may use any tricks in the book to find that formula. 
But, once you've found that formula for $\delta=\delta(\epsilon)$, you still have one more job: You must prove the statement "If $|x-a|<\delta$ then $|f(x)-L| < \epsilon$.
Finding the formula for $\delta=\delta(\epsilon)$ can sometimes be done by "solving the inequality $|f(x)-L| < \epsilon$ for $|x-a|$". And that works for the example in your question. As I solve the inequality, I will very carefully record the directions of implication, which for this problem are all bidirectional implications, i.e. they are all "if and only if"s.
\begin{align*}
&|(2x-5)-3| < \epsilon \\
\iff &|2x - 8| < \epsilon \\
\iff &2|x-4| < \epsilon \\
\iff &|x-4| < \frac{\epsilon}{2}
\end{align*}
Based on this work, I have found the appropriate formula:

Let $\delta = \frac{\epsilon}{2}$

And now the last part: I must prove the statement "If $|x-4| < \delta$ then $|(2x-5)-3| < \epsilon$". 
So, assume
$$|x-4| < \delta
$$
and substitute $\delta=\frac{\epsilon}{2}$ to get
$$|x-4| < \frac{\epsilon}{2}
$$
And now you can see why I was sooooo careful to record bidirectional implication arrows, i.e. all those $\iff$ arrows, because I can just follow the implications backwards from $|x-4| < \epsilon/2$ to $|(2x-5)-3| < \epsilon$.
A: The definition of limit ask to prove that for any $$\varepsilon >0$$ there exists a $$\delta(\varepsilon)$$ such that if $$0<|x-x_0|< \delta$$ then $$|f(x)-l|<\varepsilon$$
In your example, let be $$\varepsilon >0$$ and consider the inequality
$$|(2x-5)-3|< \varepsilon$$. You get
$$|2x-8|<\varepsilon$$ which is satisfied for $$|x-4|<\frac{\varepsilon}{2}$$. If you put $$\delta(\varepsilon)=\frac{\varepsilon}{2}$$ you have done. Therefore there is no circular reasoning.
