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Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials

$$\displaystyle \lim_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}.$$

I do not know where to start on this, I tried multiplying it out and that didn't help really. It seems very complicated and I know I have to reduce it somehow but everything I do just makes it more complicated.

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marked as duplicate by Qiaochu Yuan Sep 7 '11 at 19:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Jordan, Are you able to calculate the limit $\lim_{x \to \infty} \frac{x^2+1}{x^2+x}$? If you can, then this is pretty much the same idea... – Srivatsan Sep 6 '11 at 19:21
Most likely not, I am terrible at math and I do not seem to be able to retain anything I learn, ever. It should be zero i think though just because it is 1/infinity. – user138246 Sep 6 '11 at 19:23
Well, the answer to that is not zero. What do you think about $\lim_{x \to \infty} \frac{x}{x+1}$? Also, what about $\lim_{x \to \infty} \frac{x+1}{x}$? Don't worry; I am asking only because having an idea about how much you understand is very helpful for a user answering your question... – Srivatsan Sep 6 '11 at 19:26
It should be 1. – user138246 Sep 6 '11 at 19:29
@Jordan I made a flowchart for evaluating limits when I taught Calculus last semester. I hope it is helpful for you. Let me know if you need any help reading it, since it's a little bit terse. – Austin Mohr Sep 6 '11 at 19:36
up vote 6 down vote accepted

Hint: Divide the numerator and the denominator by the highest power of $x$ that occurs (in this case, $x^4$); distribute it so that it shows you exactly what is going on. For example, $$\begin{align*} \frac{1}{x^4}(x-1)^2(x^2+x) &=\frac{1}{x^2}\times\frac{1}{x^2}\times(x-1)^2\times(x^2+x) &&\text{(factor }\frac{1}{x^4}\text{)}\\ &= \frac{1}{x^2}\times(x-1)^2 \times \frac{1}{x^2}\times (x^2+x) &&\text{(reordering)}\\ &= \left(\frac{1}{x^2}(x-1)^2\right)\times\left(\frac{1}{x^2}(x^2+x)\right)&&\text{(associativity)}\\ &= \left(\frac{1}{x}(x-1)\frac{1}{x}(x-1)\right)\times\left(\frac{1}{x^2}(x^2+x)\right)\\ &=\left(\frac{x-1}{x}\times\frac{x-1}{x}\right)\times\left(\frac{x^2+x}{x^2}\right)\\ &=\left(\frac{x-1}{x}\right)^2\left(\frac{x^2+x}{x^2}\right). \end{align*}$$ Now simplify a bit. Then do the same thing with the numerator, and see what happens as $x\to\infty$.

(Once you have more familiarity with end behavior and limits of rational functions, you'll be able to compute this limit "by eye").

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I am not quite following the math you used. Why does diving (x-1)^2 by x^4 give you 1/x^2 (x-1)^2? – user138246 Sep 6 '11 at 19:22
@Jordan: It doesn't: You use two of the $x$s in $x^4$ for $(x-1)^2$, and the other two $x$s for the $(x^2+x)$. In other words, first write $\frac{1}{x^4}$ as $\frac{1}{x^2}\times\frac{1}{x^2}$, then use one factor for the $(x-1)^2$, and the other factor for $(x^2+x)$. – Arturo Magidin Sep 6 '11 at 19:25
I get it, it just confused me because I never see it written like that I guess. Not sure why I couldn't think of that myself. Another thing I don't understand is why it came out to be (x-1)^2 /x shouldn't it be x-1/x^2? – user138246 Sep 6 '11 at 19:28
And thanks to the downvoter for pointing out what the confusing part might be. – Arturo Magidin Sep 6 '11 at 20:00
@Jordan: You need your algebra skills to be up to date to do calculus; maybe you just need a refresher, but you do need them. – Arturo Magidin Sep 6 '11 at 20:10

Imagine multiplying out the top, and the bottom (imagining is less work than doing). The top is $4x^4$ plus some lower degree junk. The bottom is $x^4$ plus some lower degree junk. The behaviour of a polynomial for large $x$ is determined by its term of highest degree. So for large $x$, our expression should be about $4x^4/x^4$, that is, about $4$.

The above reasoning is the heart of the matter. However, it is likely that you are expected to give a more formally justified answer. There is a standard trick that makes the above reasoning more precise.

Take $x \ne 0$, and divide the top by $x^4$. We get $(2+ 1/x^2)^2$. Divide the bottom by $x^4$. We get $(1-1/x)^2(1+1/x)$. So for $x\ne 0$, our expression is equal to $$\frac{\left(2+\frac{1}{x^2}\right)^2}{\left(1-\frac{1}{x}\right)^2\left(1+\frac{1}{x}\right)}.$$

Now we can see exactly what happens as $x\to\infty$.

Comment: Multiplying out top and bottom, as you did, also gives a very good start. The next step then is to divide top and bottom by $x^4$. So for example when you multiply out the top you get $4x^4+4x^2+1$. When you multiply out the bottom you get $x^4-x^3-x^2+x$. So we are interested in $$\lim_{x\to\infty}\frac{4x^4+4x^2+1}{x^4-x^3-x^2+x}.$$ Now divide top and bottom by $x^4$. We want $$\lim_{x\to\infty}\frac{4+\frac{4}{x^2}+\frac{1}{x^4}}{1 -\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}}.$$ The answer is now obvious.

That is a perfectly fine approach. Some people may consider it harder than the first solution that I gave. However, you may well consider it easier. The only thing that bothers me about it is calculating all those lower degree terms, all the while knowing they won't make a bit of difference!

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By substitution.

I think the OP is having some difficulty in dividing by a power of $x$ and simplifying the rational function, in terms of $1/x$. I must admit that the simplification process seems a bit unnatural while working with $1/x$.

So, my suggestion is that, at the very beginning, substitute $y = 1/x$. Then $x \to \infty$ is equivalent to $y \to 0$. (Note that we should be talking about the one-sided limit as $x \to 0^+$, but we will ignore that issue here.)

Plugging this in the numerator, we get: $$ (2x^2+1)^2 = \left(2\frac{1}{y^2} + 1\right)^2 = \left(\frac{2+y^2}{y^2}\right)^2 = \frac{(2+y^2)^2}{y^4}. $$ Similarly the denominator becomes: $$ (x-1)^2(x^2+x) = \left( \frac{1}{y} - 1\right)^2 \left( \frac{1}{y^2} + \frac{1}{y} \right) = \frac{(1-y)^2}{y^2} \times \frac{1+y}{y^2} = \frac{(1-y)^2(1+y)}{y^4}. $$

Dividing the numerator by the denominator, we get $$ \frac{(2x^2+1)^2}{(x-1)^2(x^2+x)} = \frac{(2+y^2)^2}{y^4} \times \frac{y^4}{(1-y)^2(1+y)} = \frac{(2+y^2)^2}{(1-y)^2(1+y)}. $$ (Notice that the problematic $y^4$ terms nicely cancel.)

Finally, we have $$ \lim_{x \to \infty} \frac{(2x^2+1)^2}{(x-1)^2(x^2+x)} = \lim_{y \to 0} \frac{(2+y^2)^2}{(1-y)^2(1+y)}. $$ I believe that the latter limit should be easy to solve (essentially by plugging in $0$).

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For rational functions and when limits are taken as $x \rightarrow \infty$ or $x \rightarrow -\infty$, the answer will be the same if you only keep the greatest degree term on top and the greatest degree term on bottom. Several of the answers already given indicate why this will be the case (once you understand things a little better), but for this reason those answers involve more work than is strictly necessary to find the limit (as opposed to justifying that what you get is in fact the limit). What I said in the first sentence will greatly reduce the work you have to do once you realize that you don't have to multiply everything out to get your hands on the terms with the greatest degrees.

$$\lim_{x \rightarrow \infty} \frac{(2x^2 + 1)^{2}}{(x-1)^{2}(x^2 + x)} = \lim_{x \rightarrow \infty} \frac{(2x^2)^{2} +\; ...}{(x^2 + \; ...)(x^2 + x)}$$

$$ = \lim_{x \rightarrow \infty} \frac{4x^2}{(x^2)(x^2)} = \lim_{x \rightarrow \infty} \frac{4}{1} = 4$$

The key in getting this to reduce your work is in learning how to efficiently get your hands on the terms with the greatest degrees (these terms are often called the dominant terms). Here's an example that better illustrates what I'm talking about than the example you had. Consider

$$ \lim_{x \rightarrow \infty} \frac{(3x^3 - 5x + 8)^{2}(2x - 1)^2}{(5 - 2x^2)^3(9x^2 - 16)}$$

Expand the numerator just enough to get your hands on the numerator's dominant term:

$$(3^{2}x^{6} +\; ...)(2^{2}x^{2} + \;...) = (9)(4)x^{8} +\; ...$$

Expand the denominator just enough to get your hands on the denominator's dominant term:

$$(-2^{3}x^{6} + \; ...)(9x^2 + \; ...) = (-8)(9)x^{8} + \; ...$$

Therefore, we get

$$ \lim_{x \rightarrow \infty} \frac{(3x^3 - 5x + 8)^{2}(2x - 1)^2}{(5 - 2x^2)^3(9x^2 - 16)}= \lim_{x \rightarrow \infty} \frac{(9)(4)x^{8}}{(-8)(9)x^{8}} = -\frac{1}{2}$$

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HINT $\displaystyle\rm\ \frac{a_4\ x^4 + a_3\ x^3 + a_2\ x^2 +a_1\ x + a_0}{b_4\ x^4 + b_3\ x^3 + b_2\ x^2 +b_1\ x + b_0}\: =\: \frac{a_4\ + \frac{a_3}x + \frac{a_2}{ x^2} +\frac{a_1}{x^3} + \frac{a_0}{x^4}}{b_4\ + \frac{b_3}x + \frac{b_2}{ x^2} +\frac{b_1}{x^3} + \frac{b_0}{x^4}}\to \frac{a_4}{b_4}\: $ as $\rm\: x\to \infty$

Scaling numerator, denominator by $\rm\:x^{-4}\:$ essentially changes variables to $\rm\ z = 1/x = 0 \ $ vs. $\rm\ x = \infty\:,\ $ reducing it to the simpler limit of a rational function at $0$. Many limits at $\rm\:x = \infty\:$ are simplified by changing variables to $\rm\:z = 1/x = 0\:.\:$ As we saw above, for rational functions, this variable change is quickly achieved by a simple scaling.

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On seeing your hint, I think perhaps explicitly substituting $z = 1/x$, simplifying the resulting rational function, and then letting $z \to 0$ might seem more intuitive to the OP. (I.e., if s/he is not comfortable with dividing by $x^{4}$...) – Srivatsan Sep 6 '11 at 19:42

When you multiply out, you are going to have a polynomial on top, and a polynomial below. First step is to figure out the degree of those polynomials. You don't have to worry about the exact coefficients. Just find the degree.

If the degree of the top polynomial is greater than that of the bottom polynomial, the limit will be $\infty$ or $-\infty$. If the degree of the top polynomial is less than that of the bottom polynomial, the limit is 0.

If the degree is the same, say $n$, then divide both the top and bottom by $x^n$. That reduces both the top and the bottom to the form $a+b/x+c/x^2+...$, and each of those is easy to evaluate as $x$ goes to $\infty$.

Now once you see that it is going to be in that form, you can notice that you don't have to bother multiplying it out to actually get the polynomials. You should be able to figure out what the degree will be without actually multiplying them out completely. Once you have the degree, say $n$, you can do the division of numerator and demoninator by $x^n$ as they stand, just distributing the $n$ $x's$ among the factors according to their degrees.

I don't want to fully solve your example since it is homework, but just to get you started, the degree is 4. The two factors in the top each get divided by $x^2$ as do each of the two factors in the bottom. The $(x^2+x)$ factor would then become $(1+1/x)$.

When you do that for all the factors, each factor will end up being something that goes to a constant as $x$ goes to $\infty$, and you should be able to see the limit.

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