$\epsilon - \delta$ limit evaluation logic The definition is as follows; $$\forall \epsilon > 0, \exists \delta >0 : 0 < |x-a| < \delta \implies |f(x) - L| < \epsilon$$
Now I understand this definition, well I thought I did, until I encountered an example.
Now I have been told that to check that a limit $L$ is correct(known limit), then I must find a $\delta$ that satisfies $ 0 < |x-a| < \delta$ which then  $\implies |f(x) - L| < \epsilon$.
Now looking at a few examples, they start off with $|f(x) - L| < \epsilon$, manipulate it until they find a $\delta$ which works. But this is then $|f(x) - L| < \epsilon \implies 0 < |x-a| < \delta$, but don't we want the implication the other way around?
 A: Glad that you asked this question. The definition of limit does not say that $$|f(x) - L | < \epsilon \Rightarrow 0 < |x - a| < \delta$$ but rather the reverse implication is desired. Then why do most $\epsilon-\delta$ proofs start with $|f(x) - L| < \epsilon$ and manipulate it in a crafty manner to yield something like $0 < |x - a| < \delta$?
Well, for some specific (and simpler) limit problems like $\lim_{x \to 2}2x = 4$ the implication works both ways so that $\Leftrightarrow$ kind of implication is valid and only in these kinds of problems we start with inequality $|f(x) - L| < \epsilon$ and manipulate it to get $0 < |x - a| < \delta$.
For somewhat complex limit like $\lim_{x \to 1}x^{2} = 1$ we need to work exactly according to the definition. We start with $\epsilon > 0$ and fix the goal to achieve as $$|x^{2} - 1| < \epsilon\tag{1}$$ The means to achieve the goal is to bound the difference $|x - 1|$ by a suitable number $\delta$ such that the goal $(1)$ is achieved. To achieve our desired goal it is important to analyze our goal a bit and develop a certain strategy which leads from our means to our goal. The goal can be written $$|x - 1||x + 1| < \epsilon$$ and if we can somehow bound this factor $|x + 1|$ by some number say $K$ the goal $(1)$ is reduced to a simpler goal $$|x - 1| < \frac{\epsilon}{K}\tag{2}$$ and it is sufficient to ensure $(2)$ in order to fulfill $(1)$.
But this simplicity comes at a price of bounding the factor $|x + 1|$ by some suitable $K$. Now we take help of our intuition and note that since $x \to 1$ we are dealing with values of $x$ near $1$ and hence we may suppose $x$ to be positive and less than $2$ i.e. we fix $|x - 1| < 1$. Doing so bounds the factor $|x + 1|$ by $3$ so that we can take $K = 3$ and our goal $(2)$ is now $|x - 1| < \epsilon/3$. It is now clear that if we chose $\delta$ as minimum of $1$ and $\epsilon / 3$ then we achieve two things if $0 < |x - 1| < \delta$: first we bound the factor $|x + 1|$ by $3$ and second we ensure $|x - 1| < \epsilon/3$. Together these ensure the original goal $|x^{2} - 1| < \epsilon$ and $\delta = \min(1, \epsilon/3)$ works.
Note that the definition of a limit does not specifically ask you to find $\delta$ in terms of $\epsilon$ explicitly like $\delta = \min(1, \epsilon/3)$. It only asks you to ensure that for every $\epsilon > 0$ there is some $\delta > 0$ which leads to the desired implication. Moreover it is also obvious that if some $\delta$ works for a given $\epsilon$ then any $\delta' < \delta$ also works for that same $\epsilon$ and hence the definition does not require any unique $\delta$ for a given $\epsilon$. Thus for a typical $\epsilon-\delta$ proof question in exam, the answers of two students may vary significantly.

Another popular strategy (apart from factoring and bounding one factor) is to split the term $|f(x) - L|$ into multiple parts (say into $n$ parts) and ensure that each part is less than $\epsilon / n$ (think of divide and conquer). This breaks our final goal into a number of smaller goals and each smaller goal needs to be analyzed further to find suitable $\delta$'s for each smaller goal. The minimum of all such $\delta$'s works to ensure the final goal.
A: It's very good that you take note of what direction you should and should not work in; however, there is more to it than that. This is actually a proof technique that works by forming a chain of double implications between something you want to prove that's hard and something easier to prove. 
When you're trying to prove something, you can start with what you're trying to prove to look for "if and only if's" $(\leftrightarrow$'s) to generate equivalent statements to the first one until you find something that you know you can prove or know is already true. It might go 3 or 4 lines  but once you reach something you can prove, you start the proof by proving the last line of your "if and only if's" chain and working in reverse. 
It's a very basic and useful proof technique. Here is how it works: Suppose you want to prove A; but you don't know how to do it. After toying around with A on your scratch paper, you determine A is logically equivalent to B, which is logically equivalent to C,....and so forth: $A \leftrightarrow B\leftrightarrow C \leftrightarrow D$
You realize that you know D is a true condition; or that you can at least easily prove it is a true condition. When working like this, that's the point where I say BINGO! Time to start the proof! Then I pull out a fresh sheet of paper, and I prove D. Once D is true, I know D implies C and C implies B and B implies A. To write my proof, I literally just copy down the lines I wrote on my scratch paper in reverse ending with A. 
The presentation of the proof will make it look as if you just randomly thought of D, out of nowhere, even though it had no visibly obvious relation to A, and then proved A based on it, making you look like some sort of genius who can see things that others can't.
I'd advise you learn this basic and extremely useful logical technique ASAP.
Warning: This is NOT the only tool you should have in your box. It should be one of many. There are lots of areas of theory where there are not really that many if and only if statements that can be made, and then you'll have to start thinking outside of the box. In other words, don't make the mistake of using a shovel to do a proof in a case when you should be using a scalpel or a lock pick. 
-Adam
