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?
 A: 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.
A: Multplying by the conjugate quantity does not take away the discontinuity ($2$ is still a root of the denominator $x^2-4$) at $x=2$. What you can say is :
$$ \lim \limits_{\substack{x \to 2 \\[1mm] x < 2}} \frac{x+2}{x-2} =  - \infty $$
and
$$ \lim \limits_{\substack{x \to 2 \\[1mm] x > 2}} \frac{x+3}{x-2} = + \infty. $$
Edit : The example you gave when you edited you post could be generalized : let $P$ and $Q$ be two polynomial functions defined on $\mathbb{R}$ and $a \in \mathbb{R}$ such that : $a$ is a root of $P$ and $Q$ with multiplicity $1$. Since $a$ has multiplicity $1$, there exist a polynomial function $P_{1}$ (resp. $Q_{1}$) such that $\forall x \in \mathbb{R}, \, P(x) = (x-a)P_{1}(x)$ and $P_{1}(a) \neq 0$ (resp. $\forall x \in \mathbb{R}, \, Q(x) = (x-a)Q_{1}(x)$ and $Q_{1}(a) \neq 0$). The function 
$$ x \, \longmapsto \, \frac{P(x)}{Q(x)} $$
is not defined at $a=0$ but note that :
$$
\begin{align*} 
\frac{P(x)}{Q(x)} &= {} \frac{ \require{cancel} \cancel{\color{blue}{(x-a)}}P_{1}(x) }{ \require{cancel} \cancel{\color{blue}{(x-a)}}Q_{1}(x) } \\[2mm]
 &= \frac{P_{1}(x)}{Q_{1}(x)} \\
\end{align*}$$
As a consequence :
$$ \lim \limits_{x \to a} \frac{P(x)}{Q(x)} = \lim \limits_{x \to a} \frac{P_{1}(x)}{Q_{1}(x)} = \frac{P_{1}(a)}{Q_{1}(a)}. $$
In your example : $a=-2$, $P(x) = x+2$, $P_{1}(x) = 1$ and $Q(x) = x^{3} + 8$, $Q_{1}(x) = x^{2}-2x+4$. 
A: 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 $$\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to2} \frac{1}{x^2-2x+4} = \frac{1}{12}. $$
A: Answer to your edited question:
The functions $f(x)=\dfrac{x+2}{x^3+8}$ and $g(x)=\dfrac{1}{x^2-2x+4}$ are not equivalent, since $x=-2$ is not in the domain of $f$, but is in the domain of $g$.
When you're looking for the limit, you are interested in the behaviour as $x$ approaches $-2$, and both functions have the same behaviour in this case. That's why it's allowed to algebraically manipulate the rational function as far as limits are concerned.
