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?

The mistake you are making is that $\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to -2}\frac{x+2}{(x+2)(x^2-2x+4)} = \frac{0}{0}$ is not necessarily discontinuous. It is a so called indeterminant form. Typically, you have to do something algebraic to "get to" the actual limit, similar to what you did by factoring and cancelling. So indeed $$\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to2} \frac{1}{x^2-2x+4} = \frac{1}{12}. $$
