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?

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Since both the numerator and denominator are degree 2 polynomials, just factor $x^2$ out of each, cancel (which you can do--since $x\rightarrow -\infty$, you may as well assume that $x<0$), and see what happens to what's left as $x\rightarrow -\infty$. – user5137 Nov 22 '11 at 0:06
It is certainly allowable to (in effect) define a new variable $y=-x$ and make the replacement. Whether you think that simplifies things is up to your taste. I don't think it does. – Ross Millikan Nov 22 '11 at 0:23

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.

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You solved this limit in the same way as $x \to \infty$,that's why im wondering how to solve,with minus or plus infinity, or depends of the function? – nEAnnam Nov 22 '11 at 0:16
@nEAnnam In your particular example, it did not matter. I just gave an example at the beginning to show that it could matter. I will try to expand on this. – process91 Nov 22 '11 at 0:18
I understand right now. Thank you very much. – nEAnnam Nov 22 '11 at 0:37

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.

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