Can someone explain the actual use of idea of limits in layman terms for me as an absolute beginner in calculus? As a beginner in calculus I have always struggled I the area of limits not when I solve higher order thinking questions but just getting the basic idea and the notion of finding limits for a function. It would be a great relief if someone could help me with this query ?
 A: I disagree with the comments.
Further, since this is an interpretation question, I feel justified in providing an answer even though the OP (i.e. original poster) has shown no work.

I will explain the notion of limits in the simplified world of single variable functions, where both the domain and range of the function is some subset of the Real Numbers.  This should give you a reasonable intuitive grasp of the idea behind the limit.
Then, you will have to broaden your intuition to consider functions that have other domains or other ranges.

The first concept to consider is the notion of a neighborhood.  The simplest example is to consider a fixed value $a \in \Bbb{R}$.  Then, for a small positive value $\delta$, the neighborhood of radius $\delta$ around the value $a$ is regarded as the set of all $x \in \Bbb{R}$ such that $-\delta < (x-a) < \delta.$
Typically, the shorthand expression for this is $|x-a| < \delta.$  Typically, in the definition of a limit, one is concerned with those values of $x$ that are in the neighborhood of radius $\delta$ around $a$, but where $x \neq a.$
Typically, this is expressed as $0 < |x-a| < \delta.$

Then, you have to understand the idea of (for a specific $\epsilon > 0$) the neighborhood of radius $\epsilon$ around some fixed finite value $L$.
Basically, this neighborhood is expressed as the set of all $y$, such that $|y - L| < \epsilon.$

Now, you are ready for the intuitive definition of a limit.
Suppose that you see the assertion that 
$\displaystyle \lim_{x \to a} f(x) = L$.
Assigning the variable $y$ to represent $f(x)$, what this assertion signfies, is that for any $\epsilon > 0$ there exists a $\delta > 0$ such that
If $x$ is in a neighborhood of radius $\delta$ around $a$, and $x \neq a$,
Then $y = f(x)$ is in a neighborhood of radius $\epsilon$ around $L$.
More formally, the assertion is written:
$\displaystyle \lim_{x \to a} f(x) = L$ signifies that
For all $\epsilon > 0$, there exists a $\delta > 0$ (where the choice of $\delta$ often depends on the choice of $\epsilon)$
such that $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon.$

As a very simple concrete example, suppose that $f(x) = 2x$, and you are asked to prove that
$\displaystyle \lim_{x\to 2} f(x) = 4.$
It turns out that for this particular problem, you can specify $\displaystyle \delta = \frac{\epsilon}{2}.$
Then, if $\displaystyle 0 < |x - 2| < \delta = \frac{\epsilon}{2}$ then you can conclude that
$|f(x) - 4| = |2x - 4| = 2|x - 2| < 2\delta = \epsilon.$
This constitutes a proof that $\displaystyle \lim_{x \to 2} f(x) = 4 ~: ~f(x) = 2x.$
The foundation of the proof was that you were able to identify a relationship between $\delta$ and $\epsilon ~\left(\text{i.e. that} ~\displaystyle \delta = \frac{\epsilon}{2}\right)$ that allowed the required constraint to be satisfied.
