Calculating limits using the $\epsilon$-$\delta$ definition. Suppose you have a function $f(x)=\frac{x^2-4}{x-2}$.
How then do we find the limit as $x\to2$ in accordance with the $\epsilon-\delta$ definition? I mean suppose we don't know how to calculate the limit and we have to derive a method to calculate the limit using $\epsilon-\delta$ definition of limit. Then what intuition will be used to derive it and what will be the value?
 A: We should have $$\epsilon>\left|\frac{x^2-4}{x-2}-l\right|=\left|x+2-l\right|,$$
for some $\delta>0$ and any $\delta>|x-2|>0$.
This is equivalent to say that the set of solutions of this last inequality must be contained in the set of solutions of the former for each $\epsilon$ and some $\delta=\delta_\epsilon$. So, we just need to solve both inequalities and impose this condition.
So we get $$\epsilon-2+l>x>l-2-\epsilon,$$ as the solution of the first inequality. 
And  $\delta+2>x>2-\delta$, $x\neq0$, for the other inequality.
Forcing that one set is inside the other we get
$$[\epsilon+l-2,l-2-\epsilon]\supset[\delta+2,2-\delta]$$ 
Since this is for any $\epsilon$ (well, it is enough to impose it for $\epsilon=1/n$, for $n\in\mathbb{N}$), and some $\delta=\delta_\epsilon$ we need to find the intersection of all the intervals $[\frac{1}{n}+l-2,l-2-\frac{1}{n}]$. This is easy in this case. It is the number $l-2$. In fact, notice $l-2$ is inside all these intervals and that if we take any other number, $l-2+t$, we can find some of those intervals in which $l-2+t$ is outside by taking $n>|1/t|$. 
Then, since the corresponding intervals $[\delta_n+2,2-\delta_n]$ should be inside the $[l-2+\frac{1}{n},l-2-\frac{1}{n}]$, and the intersections of the former for all $\delta_n$, is $2$, then the intersections must coincide. 
We get then $l-2=2$, from where $l=4$.
In general:
For the limit $$\lim_{x\rightarrow a}f(x)$$
One must solve the inequality $\epsilon>|f(x)-l|$ and the inequality $\delta>|x-a|>0$ (separately) and impose the set of solutions of the latter is inside the set of solutions of the former. Then impose this condition for every $\epsilon>0$ and some $\delta=\delta_\epsilon$ (it usually involves intersecting many intervals and it is enough to do it for $\epsilon=1/n$, for $n\in\mathbb{N}$).
A: Let $\,\epsilon >0\,$ be arbitrary. Since $\,x\to 2\,$ we are very close to $\,2\,$ but not equal to $\,2\,$, and then
$$\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2\;,\;\;x\ne 2$$
and from here that the wanted limit is $\,4\;$ . Thus:
$$\left|\frac{x^2-4}{x-2}-4\right|=|x+2-4|=|x-2|$$
then, our choice for $\;\delta\;$ is pretty simple: choose $\,|x-2|<\delta:=\epsilon\;$ and we're done...
A: To answer the OP's question: "Then please can you explain what other method can be sed to compute limit effectively and why that method works and how does it give correct answers?", since he included "infinitesimals" among the tags, I would elaborate on @Daniel Fischer's comment above and set $x=2+h$ where $h\not=0$ is infinitesimal.  Then the limit can be calculated "effectively" (or more precisely "directly") as follows:
$$\frac{x^2-4}{x-2}=x+2,$$
therefore
$$\lim \frac{x^2-4}{x-2}=\lim x+2=\text{st}(2+h+2)=4+\text{st}(h)=4+0=4,$$
since by definition of an infinitesimal the standard part "st" satisfies $\text{st}(h)=0$.  Here one does not need to know what happens for every epsilon, and needn't bother with choosing a delta, in accordance with the OP's request "Then what intuition will be used to derive it and what will be the value ?"
