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I am learning about limits and there is something that I cant quite understand:

If we have the function:

$$ f(x) =\frac{x^2-1}{x-1} $$

Let's say that we want to see which value for y (image) the function approaches as x (domain) gets closer to 1. On a nutshell, we have to take this following limit:

$$ \lim_{x\to1}\frac{x^2-1}{x-1} $$

As soon as we look to this function, we realize that the function is not continuous at x = 1 (By the way, can I say that?).

I know the algorithm to figure out the solution of the limit:

First, there is the need of eliminating the function discontinuity. Usually, it is just a matter of factorizing the function into a new function which the exactly same image as the one before with one crucial difference: The function is continuous for all real numbers

My doubts:Is my way to think about it correct? Can I think like that?

Take the example above:

$$ f(x) = \frac{x^2-1}{x-1} $$

After factorizing, we get:

$$ f(x) = {x+1} $$

If we plot both functions, they are the same, although the second one has its continuity all along the real numbers domain

Thanks in advance

Answers