Calculus of Variations. Understand Epsilon's Need? What purpose is Epsilon in Calculus of Variations?
My question being what is the need for Epsilon in the equation 
$y(x)+\epsilon n(x)$ when we assume this as the neighborhood curve of $y(x)$?
When we try to find out the Euler Lagrange condition, we take $y(x)$ to be the curve and $y(x)+\epsilon n(x)$ to be the neighboring curve.  
 A: Take a look at the derivation of the Euler-Lagrange equation, e.g. at Wikipedia. The ordinary derivative with respect to $\epsilon$ is considered. Thus, the $\epsilon$ serves to reduce the concept of variation with respect to a function to the simpler concept of differentiation with respect to a variable.
A: For a function $f$ of a real variable $t$ a small increment at $p$ is just a real number $\epsilon$ with $|\epsilon|\ll1$. In order to find conditions  that permit a local extremum of $f$ at $p$ we therefore have to study
$$f(p+\epsilon)-f(p)=f'(p)\epsilon+o(\epsilon)\qquad(\epsilon\to0)$$
and find that $f'(p)=0$ is a necessary condition (This is Calculus 101). Note that, up to sign, there is just one possible direction for a small increment.
Now in an $n$-dimensional, or even $\infty$-dimensional environment a "tiny increment" at $p$ is best described as $\epsilon u$, whereby $|\epsilon|\ll1$, and $u$ is an arbitrary fixed vector making sense in said environment. We then look at the  increments $f(p+\epsilon u)-f(p)$ in direction $u$. If $p\in{\mathbb R}^n$ then, according to multivariable calculus, 
$$f(p+\epsilon u)-f(p)=\epsilon\> \nabla f(p)\cdot u+o(\epsilon)\qquad(\epsilon\to0)\ .$$
When $|\epsilon|$ is small this will assume values of both signs for certain $u$ unless $\nabla f(p)=0$. It follows that $\nabla f(p)=0$ is a necessary condition for a local extremum of $f$ at $p$.
Similar arguments apply if $f(x)$ denotes a certain integral involving the function $x:\>[a,b]\to{\mathbb R}$. In this case both $p$ and $u$ are  functions defined on $[a,b]$. The study of such a situation then leads to the Euler-Lagrange condition: In the end all "directional derivatives" at $p$ have to be zero whatever the "increment direction" $u$.
