Basic real analysis question on the epsilon-delta format Oftentimes when you hear about "how to prove" that a specific sequence or function is converging to a specific value, people will refer to the epsilon-delta style. While I have no question about the validity of such a style, I do not see that way eye to eye as intuitive or clean. 
For example, we know that the sequence $\frac{n}{n^2+1}$ converges to 0 when $n$ gets very large. But when asked to formally prove this epsilon-delta style, I draw a blank and refer immediately to a procedural technique without really understanding the problem or the procedure. I can start with : "Let $\epsilon > 0$, then there is an $n$ s.t $n > N$. We want to show that $\frac{n}{n^2+1} < \epsilon$." 
My question here is why do we want to find an $n > N$ and how to find that $N$? (I know that $\frac{n}{n^2+1} < n$) And after I found this $N$ and use the epsilon-delta format what does this argument even mean? 
 A: Let's denote your sequence by $a_n$ so if we prove that for every $\epsilon $ we find  $N$ and if $n> N$ then $|a_n|<\epsilon$ this means that all the terms of this sequence from the term $a_N$ are close to $0$ with distance lower than $\epsilon $ so the terms accumulate at zero and this explains the definition of the limit.
A: Your instinct to begin the problem by precisely stating the definition of the limit is good, but you misstated it.  What you want is:

Let $\epsilon > 0$.  We want to find $N > 0$ such that if $n > N$, then $\dfrac{n}{n^2 + 1} < \epsilon$.

Note that in your original statement you hadn't even defined $N$ before putting it to use, and you also said "there exists $n > N$", which seems to state that some $n$ exists larger than $N$ with some property, which is different from the correct condition that all $n > N$ have that property.
Anyway, once you've sorted out the language you have only to figure out how large to make $n$ such that $n/(n^2 + 1)$ is sufficiently small.  This is not usually just a matter of manipulating the fraction, but also applying some simplifying inequalities.  For example,
$$\frac{n}{n^2 + 1} < \frac{n}{n^2} = \frac{1}{n},$$
so it's enough to find $N$ such that for $n > N$, we have $1/n < \epsilon$.  This latter inequality is equivalent to $n > 1/\epsilon$, so we can take $N$ to be any integer larger than $1/\epsilon$.
This is the "discovery" version of the proof.  The "direct" formulation just avoids the process of figuring out $N$ and uses the correct value from the start.  That is: you say that given $\epsilon > 0$, define $N$ such that $N \geq 1/\epsilon$, so that if $n > N$, we have
$$\frac{n}{n^2 + 1} < \frac{n}{n^2} = \frac{1}{n} < \frac{1}{N} \leq \frac{1}{1/\epsilon} = \epsilon,$$
as desired.
