Tina Fey not only wrote the movie, but appears in it giving a rather flattering portrayal of a caring, smart and talented (though personally troubled) math teacher. The protagonist is supposed to be quite good at mathematics, and even comments that she likes and understands mathematics because it is the same in every country. (I didn't mention, but she was home schooled in Africa where her parents worked as zoologists.) However, she is more interested in boys and popularity than in math. So, she repeatedly ignores requests to be on the school's math team. (One of the "mathletes" appears in a school talent show, singing a rap that is sexually explicit and mentions just a bit about math, if I heard it correctly.) Moreover, as an excuse to flirt with the guy sitting in front of her, she pretends not to be able to do math and even fails some tests. In the end, she starts being more true to herself and one highpoint of the film is when Lohan joins the mathletes and correctly answers a question about limits during a "college bowl"-type competition. (I wish it was more clear, however, that she really knew the answer. It looked to me more as if she made a lucky guess.)

Brief Discussion about the Limit Computation

The opposing team gives the incorrect answer -1. Lohan's team is then declared the winners after correctly stating that the limit does not exist. That the limit does not exist is really not immediately obvious because the numerator and denominator both go to zero as x approaches zero, a situation that does not by itself tell you anything. However, L'Hopital's Rule tells us that we can differentiate the numerator and denominator separately and the if the new ratio has a limit then that is also the limit of the original ratio. That is, the limit should be the same as

Comments about the Film

Mike, I do not think it is possible to determine for certain that the limit does not exist using the material you already learned. (You would need either L'Hospital's or Taylor series.) But, you could get pretty good evidence, as I'll explain below in the "technical details" section.

The Limit Computation -- Technical Details

1. A vocal minority of mathematicians dislike L'Hospital's rule and argue that any question answered using this method (which was not discovered, but essentially purchased by L'Hospital!) would be better answered using power series expansion.
2. There seems to be some disagreement about the precise definition of what it means for a limit to fail to exist. It is not particularly unusual for there to be slightly different, competing definitions in mathematics, which is why it is always wise to state the precise definition being considered. The limit above does fail to exist according to the definition commonly used in American calculus textbooks, and under most other common definitions as well.
3. Since it was not necessary to actually prove in the film that the limit did not exist, it would be sufficient to simply collect strong (but not decisive) evidence to suggest that this is the case.
4. A minor, but entirely valid, complaint is that the film misuses the word "equation".
5. A "squeeze theorem" method which allows you to determine that this limit does not exist using algebra, trig, and knowledge of some other basic limits but no differentiation.
   and
   
6. Some people see an entirely different question in the film than the one I have discussed above, which would certainly change everything!

1. Using Series Expansion: I received an e-mail from Nicolas Deslandes in France who prefers to avoid using L'Hospital's rule in this problem since (as he correctly points out) there are situations in which the limit of f'/g' fails to exist even when the limit of f/g is in an indeterminate form and exists. Although I think the argument above with one sided limits is valid, I will reproduce Nicolas' method here since I do know some mathematicians who believe that L'Hospital's rule should never be used as this Taylor Expansion method is more powerful.

   Contributed by NicolasDeslandes The simplest way for me to solve the equation was using Taylor developments and equivalents: ln(1-x) = -x + o(x) sin(x) = x + o(x) so ln(1-x) — sin(x) ~ -2x and 1-cos²(x) = sin²(x) ~ x² so (ln(1-x) — sin(x))/( 1-cos²(x)) ~ -2/x, which doesn't have a limit when x approaches 0.

   An another (anonymous) site visitor made a similar suggestion:

   Contributed by Anonymous I just finished watching the movie with my Fiance and less than a minute into the credits I found myself on Google desperately searching for the equation given in the "math bowl". I was very happy to find this site. When given a question on limits, I prefer to work with taylor expansions if possible. This yields a much more precise examination than L'Hospitals rule as this rule basically uses the 2nd coefficient in the Taylor expansion for its analysis. Thus near zero, ln (1-x) ~ -x(.5x + 1) sinx ~ x 1 - (cosx)^2 = (sinx)^2 ~ x^2 Using these approximations, the equation reduces to: -.5 - (2/x). This Limit does NOT exist b/c the value of the function approaches negative infinity from 0+ and positive infinity from 0-

2. What does it mean to say a limit does not exist, anyway? Between different countries, between different disciplines of mathematics, and perhaps between different teachers, there are differences in whether one says that a limit "exists" in the case when it is infinite. For instance, in calculus classes in the USA we generally teach our students that having an infinite limit is one particular way in which the limit can fail to exist. However, when I wrote that here I received complaints from many people who consider it differently. They consider a limit of a real function which goes to positive infinity from both sides to be a limit which exists, but is infinite. On the other hand, they might say that the limit fails to exist if the situation is as above, where the limits from the left and right are both infinite but with different signs. To further complicate things, in Complex Analysis where one works on the Riemann Sphere one would say that an infinite limit is a kind of limit that exists, but no distinction is made at all between positive and negative infinity. So, on the Riemann Sphere the limit of the function in the question exists and is infinity!

3. Ways of guessing that the limit does not exist which fall short of the rigor usually expected in a calculus class: One thing you could try to do is plug values of x that are close to zero into the function and see what happens. I tried a few values and found that when x is negative I get very big, positive numbers coming out of the function. Also, when I plug in positive values for x, I get very negative values. This is enough for me to guess that you're correct: the limit from the left is positive infinity and the limit from the right is negative infinity and so the limit itself does not exist.

   But, I really cannot be sure from what I've seen so far. I mean, it could be that if I plugged in different values for x I would see something different. No finite number of points on the graph are enough to tell me what the limit is because strange things could happen between them.

   Well, let's try to make it a bit stronger. We know that the denominator is always positive (except at x=0), so if all we're interested in is the sign then this is useful information. Also, the numerator is equal to 0 at x=0. But, what about just a bit to the left and right? If we look at the tangent line to y=ln(1-x)-sin(x) at the origin we see that it has a slope of -(1/1-0)-cos(0)=-2. In other words, it is positive on the left but negative on the right. (BTW: I'm almost using L'Hospital's rule here...this is the sort of reasoning that underlies it!)

   This again is consistent with your claim that the limit is positive infinity from the left and negative infinity from the right. However, I still have not proved it. I suppose, given what we've seen above, the limit could still be 0 from both sides (despite the numerical evidence suggesting that the values of the function are growing very rapidly as we approach 0).

   Contributed by Sam Another way to show that the limit does not exist is to rewrite the denominator 1-(cos^2(x)) as sin^2(x) by using trig identities. This gives lim x->0 (ln(1-x)-sin(x))/(sin^2(x)). This expression can be separated into two parts so it is now lim x->0 [(ln(1-x))/(sin^2(x))] - lim x->0 [1/(sin(x))]. Since the limit as x approaches 0 of [1/(sin(x))] does not exist, the limit of the entire function does not exist, just as the movie says.

   Sam, perhaps you could work out a proof based on your argument above. However, it is not as simple as you say since the first of the two limits does not exist either. Note, for example, that if f(x)=1+sin(1/x) and g(x)=sin(1/x) then f(x)-g(x)=1 for every x other than 0. Thus, the limit of f(x)-g(x) as x goes to zero is equal to 1. However, neither f(x) nor g(x) have a limit as x goes to zero. This demonstrates that lim_x->0f(x)-g(x) is not necessarily equal to lim_x->0f(x) minus lim_x->0g(x) and can converge even if neither of them do. (The point is this: If the limits of f and g existed, then you could say for certain that the limit of f-g was equal to their difference. But, knowing that the limit of f and the limit of g both fail to exist does not tell you anything about the limit of f-g.)

4. Equation?: Professor Sklar from Pacific Lutheran University wrote to say:

   Contributed by Jessica Sklar I found your site when googling info on the math in Mean Girls this evening, after telling my linear algebra students today about an error in the film. Your site is very helpful! However, it does ignore the error: the students are asked to find "the limit of [an] equation", but no equation is shown. There is an *expression*, but an equation must involve an equals sign, which is not present. I thought you might want to add a note to this on your page, since it is a major terminological error that students make (confusing equations and expressions).

   Good point! In fact, this misuse of "equation" for any mathematical expression (whether it involves an equality or not) is something I have commented on elsewhere (see The Algebraist), but I missed it here. Thanks!

5. Disagreement about what the question is in the first place! I received email from an electrical engineering major in Illinois (see below) suggesting that we were completely wrong about the limit and that it does in fact exist. After some discussion, it became clear that Greg (and others like him) are reading the question differently; where I see a "-" he sees a "*", which would certainly change things!

   Contributed by Greg I can't believe that everyone on this website is buying into the fact that the limit does not exist. I am an Electrical engineer major at University of Illinois and we were mentioning this movie the other day in one of our classes. If you are going to use L'Hopitals rule then you are not accepting that the limit does not exist and when you simplify and use the L'Hopitals rule once more, you will ironically find that the correct answer is the limit is -1. So not only was she wrong in the movie, but her opponent actually had the correct answer!!! Try it for a while and you will see that I am not incorrect.

   I do not believe there is a mathematical error in my determination of the limit using L'Hospital's rule above (nor with the same answer as determined by Taylor expansion that others have suggested). And my answer agrees with the one that is said to be correct in the film, so I have reason to think it is what the authors intended. But, there does seem to be some disagreement as to what the original question was, and that seems to explain the different answer obtained by Greg. Specifically, different viewers have a different idea as to what algebraic operation precedes the sine in the numerator of the function whose limit is being taken, as explained by Kenneth Wildenhain below:

   Contributed by Kenneth Wildenhain Dear Alex Kasman, Your analysis about the limit in Mean Girls is correct except for one note. On low definition DVDs without zooming the operation in front of the sin(x) appears to be a dot indicating multiplication, while with zooming it appears to be a minus sign. If it were a dot then the answer would indeed be -1 and Caroline Krafft (played by Clare Preuss) would have won and Cady Heron (played by Lindsay Lohan) would have lost. Caroline Krafft's problem appears to be bad glasses and a bad viewing angle rather that applying L'Hopital's rule twice incorrectly. Caroline Krafft was cheated and you can tell it in her facial expressions. This is a good example why students should show their work. It's noteworthy that the moderator does not read the limit question as he did the previous questions and thus eliminate this source of error. It is also noteworthy that unlike most entries in the telephone directory Caroline's last name ends in "FFT." Could this be a hidden reference to the Fast Fourier Transform? I doubt it. Yours truly, Kenneth Wildenhain

   Yes, it does seem that there is confusion among viewers as to whether the operation is multiplication or subtraction, which would make a big difference. I'm no longer positive that I know what was intended by the writers. Did they intend it to be unambiguously subtraction (as I always thought) so that the victory in the movie is justified? Did they intend it to be multiplication (in which case the ending of the film seems unfair)? Even if they intended us to believe that Krafft simply misread the question, this is not a very satisfying ending, in my mind, since she then has made no mathematical error and so does not deserve to lose. Do me a favor, the next time you're having lunch with Tina Fey, casually inquire about this and let me know what you find out ; )

6. A "Squeeze" Approach: Gregory Adams from Bucknell University just wrote to propose another method of determining that the limit does not exist. In this alternative method, it is not necessary to take any derivatives at all, but one must know two other limits to compare it to. He observes that by using the most famous trig identity in the form 1-cos²x=sin²x and splitting the fraction, the original expression is equal to ln(1-x)/sin²(x) - 1/sin(x) Then, since ln(1-x) is negative for x's between 0 and 1, the original expression is less than -1/sin(x) for all x's in this interval. Since we know that -1/sin(x) has the limit of -infinity as x goes to zero from the right (we do know that, don't we?) then since the original expression is even more negative it must have the same limit. Similarly, for -1bigger than -1/sin(x) for these x values...but the limit of -1/sin(x) from the left is positive infinity and since the original expression is even bigger it must also go to positive infinity as x approaches zero from the left. Once again, we can conclude that the limit does not exist under either of the standard definitions. (Thanks, Greg.)

   Contributed by Greg Adams Hi Alex, I prefer to think of this as number sense rather than a squeeze theorem. ln(1-x) and -sin(x) have the same sign, so cancellation can't make their sum smaller. Thus, | ln(1-x) - sin(x) | is no smaller than |sin(x)|. Consequently, the term sin^2(x) in the denominator can't be balanced out by the numerator. Notice that the same limit for ( ln(1-x) + sin(x) ) / (1-cos^2(x) ) does exist and here the Taylor expansion quickly gives the answer. I find that too many students want to rely on methods like L'Hospital's Rule to calculate limits and ever fewer have intuition. Perhaps growing up on a slide rule enables me to look at a rational function and immediately know which asymptotes to expect. My students don't seem to come to class with this understanding. We also don't help them by stressing the formalisms over the heuristics. Both approaches have a role to play. I suspect that those who argue for using Taylor expansions over L'Hospital's rule are really arguing for the heuristic approach and its intuitive power.