limit cal

You are not actually finding the "true" answer of the problem, but rather the boundaries of the "limit" (infinity) of the number based on the lower limit, and upper limit of your graph.

Answer

That's not true. Limits are a precisely defined concept. At least, if you know what the context is then they are. Since you mentioned calculus, I'm going to assume you're interested in the definition of a derivative.

First, some notation. For any number x, the term |x| means the absolute value of x. So |3|=3, |-5|=5, |-2.7|=2.7, |7.8|=7.8, etc.

Suppose f:R->R. Then f'(p), the derivative of f at p (if it exists) is defined as:

lim { (f(p+h)-f(p))/h }
h->0

What does this actually mean? The intuition is this: the smaller h is, the closer (f(p+h)-f(p))/h gets to the derivative.* We say (f(p+h)-f(p))/h tends to f'(p) as h tends to 0. In other words, we can make the difference
| { (f(p+h)-f(p))/h } - f'(p) |
as small as we want, just by forcing h to be small. More formally: For any positive real number e, there is a positive real d such that, for any real h with |h|

| { (f(p+h)-f(p))/h } - f'(p) | <>

The derivative f'(p) is defined as the only real number with this property. You can never have more than one number with this property. There might not be any, in which case the function f is not differentiable at p.

* The intuition here is as follows: (p+h,f(p+h)) is a point on the curve close to (p,f(p)), the point we are interested in. The line between these points (let's call it L) is almost the same as the tangent to the curve (T), and its gradient is almost the gradient of the tangent (which is the derivative). But the gradient of L is
(f(p+h)-f(p))/h
and therefore this quantity is close to f'(p).