Mean Girls isn’t really a “new” movie, and I was like in grade 9 or 8 when it came out. And was amazed at the complex math that I didn’t understand in it. But after all this time, and me learning so much about math and being quite good with limits I thought that it was probably a really easy question so I googled it to check. However, I was quite interested with this problem, shown here:
Okay, so obviously the first thing I noticed was that 1 - cos^2 x = sin^2 x, that just might simplify things, but not very useful.
Another easy thing to notice is that if we plug in “0” for x, we get 0/0 because sin 0 = 0, cos 0 = 1, and ln 1, meaning to what power we have to raise e to to get 1, being 0.
Normally people would just normally jump straight into L’Hopital’s rule which can be summarized into:
If exists and becomes 0/0 or infinity/infinity then is correct to use.
If we follow our initial gut feeling to use L’Hopital’s rule the problem becomes :
lim [x->0] (-1/(1-x) - cosx)/(-2cosxsinx) = -2/0 = inf. We can thus say that the limit is infinity or in some textbooks, doesn’t exist. Normally both would be right, however, we made a mistake here. The restriction of the first condition doesn’t come up often, but does here.
A limit only exist if . Which means that the limit of the number approaching from the negative and positive sides are equal. And in this case,
Log being another way of writing ln, or natural logarithm, depending on what country you’re in, different notation is used. The reason for this difference in side limits is because of the ln x function and sin x function respectively:
with the limit of approaching 0 slightly from the negative side, the ln(1-x) becomes a slightly positive 0 while the sin x becomes slightly negative and since it is being subtracted the entire top becomes a slightly positive 0, but with the limit approaching 0 slightly from the positive side, ln(1-x) becomes slightly negative and sin x becomes slightly positive and then is subtracted from the slightly negative ln(1-x). Thus the limit does not exist.
This question requires a little be of a different approach than most limits, so in that sense, is a really good question.
— Emil Kerimov