Limit calculation and discontinuity Having a function, which has a polynomial in the denominator like:
$$ \lim_{x \to 2}\,\dfrac{x+3}{x-2} $$
We see there is a discontinuity at x=2, because it sets the denominator to 0. But multiplying by its conjugate we get:
$$ \dfrac{x+3}{x-2}\cdot\dfrac{x-2}{x-2} = \dfrac{(x+3)(x-2)}{x^2-4} $$
And now there is no discontinuity in $x=2$ anymore, but should not that function be an equivalent of the original one? we are just multiplying by $1$.
UPDATE:
Apologies, I chose a very poor example. 
Consider this one:
$$ \lim_{x \to -2}\,\dfrac{x+2}{x^3+8} $$
When doing f(-2) you get 0/0. By doing factor, cancel and plug in: 
$$ \lim_{x \to -2}\, \dfrac{(x+2)}{(x+2)(x^2-2x+4)} $$
$$ \lim_{x \to -2}\, \dfrac{1}{x^2-2x+4} $$
Then f(-2)=1/12, it is not 0/0 anymore. So my question is that how come that two functions that should be equivalent, have different discontinuit?

What is $x^2 - 4$ at $x = 2$? You still have a denominator of zero. 
As $x \to 2$, the limit of $\dfrac {x+3}{x -2}$ does not exist.
$$\lim_{x\to 2^-} \dfrac{x+3}{x-2} = -\infty$$
$$\lim_{x\to 2^+} \dfrac{x+3}{x-2} = +\infty$$
Edit: Note that the functions are equivalent everywhere except at $x=-2$. The original (second function in the post) is not defined at $x = -2$, whereas the subsequent is defined there. Since they have different domains on which each is defined, they are not strictly equivalent. 
Recall that when finding the limit as $x\to -2$, we are not interested in the behavior of the function at $x = -2$, only as $x$ approaches $-2$, and for this purpose (finding the limiting behavior of the function as $x\to -2)$, the functions are equivalent.
