The limit of a rational function $$ \lim _{x \to -\infty} \frac{3x^2+3x}{2x^2+2}$$
Is a good practice to do this?
Change the $ -\infty $ to $ \infty $, and change the sign of the $ x $ variables:
$$ \lim _{x \to \infty} \frac{3(-x)^2+3(-x)}{2(-x)^2+2}$$
Or is the same result when $ x \to -\infty$ that $ x \to \infty$
Which way would be the right to solve this kind of limits?
 A: Your first method was correct. The limits as $x\to-\infty$ or $x\to\infty$ can produce very different results, for instance:
$$\lim_{x\to\infty} x \ne \lim_{x\to -\infty}x\ .$$
Here's how to go about solving this limit:
$$\lim_{x\to - \infty} \frac {3x^2+3x}{2x^2+2}=\lim_{x\to -\infty}\left(\frac {3x^2+3x}{2x^2+2}\cdot\frac{\frac 1 {x^2}}{\frac 1 {x^2}}\right)=\lim_{x\to - \infty} \frac {3 + \frac 3 x}{2 + \frac 2 {x^2}}=\frac 3 2$$
The last inequality follows by noting that:


*

*The limit of a quotient is the quotient of the limits

*The limit of a sum is the sum of the limits


In general, when you have $x\to\infty$ or $x\to -\infty$ and a rational function, try dividing out the highest degree of $x$ from the numerator and denominator.

In the example above, it did not matter that $x\to -\infty$ rather than $x\to\infty$, because all the terms which were dependent on $x$ approached 0 in the limit. A small change to the example, however, does yield different results:
$$\lim_{x\to- \infty}\frac {3x^3+3x}{2x^2+2}=\lim_{x\to-\infty}\frac{3x+\frac 3 x}{2+\frac 2 {x^2}}=-\infty$$
The last equality comes from the observation that the numerator of this expression will decrease without bound, while the denominator approaches 2. Suppose instead that the limit subscript had been $x\to\infty$. Then:
$$\lim_{x\to \infty}\frac {3x^3+3x}{2x^2+2}=\lim_{x\to\infty}\frac{3x+\frac 3 x}{2+\frac 2 {x^2}}=\infty$$
My point is just that you do not get the same result. One way of handling the $-\infty$ case is to make the substitution you suggest, another way would be to just think (similarly as you do for $x\to\infty$) about what happens to the function as $x$ decreases without bound.
A: Intuition about positive quantities is sometimes better than intuition about negative quantities.  In that case, a transformation of the type you mention can be useful.  If you are going to make a change of variable, I would advise letting $w=-x$, and writing
$$\lim_{x\to-\infty}\frac{3x^2+3x}{2x^2+2}=\lim_{w\to\infty} \frac{3w^2-3w}{2w^2+2}.$$
There is nothing wrong with writing $\displaystyle\lim_{x\to\infty}\frac{3x^2-3x}{2x^2+2}$, but reusing  $x$, with changed meaning, could lead to confusion in more complex situations. 
In this particular problem, there is no real gain in making the change of variable, since what is happening should be intuitively clear. For large negative $x$, the top is dominated by $3x^2$, and the bottom by $2x^2$, so the ratio will be nearly $3/2$.  The usual formal way of making sure is to divide top and bottom by $x^2$. Then the behaviour as $x$ becomes large negative is transparent.
