 This video is about a calculus problem that the great Russian mathematician of Vladimir Arnold liked to challenge other professional mathematicians with in his lectures. So he would ask them to calculate the following limit. We take sin of tan of x minus tan of sin of x and divide it by arc sin arc tan of x minus arc tan arc sin of x. So the point about his problem was that this is almost impossible to solve using the standard techniques that mathematicians use, but there's a very simple way of working out the answer with almost no effort at all. So first of all, let's see what goes wrong with the standard techniques. So your standard reaction on seeing a limit like this is to apply de L'Hopital's rule. You recall this says that the limit of x tends to zero of f of x over g of x is the limit of x tends to zero of f prime x over g prime x provided f of zero equals g of zero equals zero and f and g are nice functions. I'm not going to say exactly what nice means because I don't really care. So for example, suppose you want to calculate the limit as x tends to zero of one minus cosine x over x squared. Well, the numerator and denominator both have limit zero, so we just differentiate them and we find it's the same as the limit of sin x over two x. Well, again, the numerator and denominator both vanish at x equals zero, so we differentiate them again and we get cosine x over two. Well, this time they have none zero limit. In fact, this is limit one and the denominator rather obviously is limit two. So this limit here is just equal to two. So why don't we just apply this rule to this expression here? Why doesn't it work? Well, the answer is it does work, but there's a bit of a catch. The catch is that instead of differentiating this expression once or twice, you have to differentiate it seven times and every time you differentiate it, it gets about twice as horrible as it was before because of Leibniz's rule. So doing it seven times, the expression you get is going to be about a hundred times as large as this. You can do this on a computer algebra system. I tried it on one computer algebra system and it managed the numerator and gave up on the denominator that said the numerator, if you take its derivative seven times and take the limit, you get minus 168. So it's possible to do it by computer. Anyway, Arnold, the reason Arnold liked this example is it was kind of a good way of making people look stupid because there's actually a really easy solution. So what you notice is you're trying to find the limit as x tends to zero of f of x minus g of x over g inverse of x minus f inverse of x, where this is the inverse function of g. So if g was sine, this would be arc sine. And here we're taking f of x to be sine of tan of x and g of x is equal to tan of sine of x. So what we drew is we just draw a picture of what is going on because drawing a picture is always a really good way to understand what's going on. So this is going to be the line y equals x and I'm going to draw a blue line for the graph of g of x and a red line for the graph of y equals f of x. And the point about these functions g and f is that f prime of zero equals g prime of zero equals one. And that's pretty much all we need to know about the functions plus the fact they're not too ugly. And now what you do is you choose a point x. So this is going to be our point x. And I look at this distance here. So this distance here is just f of x minus g of x. And I can look at this distance here. And this is g inverse of x minus f inverse of x because this point x is here. So this distance here is given by that expression. So the limit we want to find out is the ratio of this line to this line. And you can see from geometry that if you're really close to the origin, then the ratio of this line to this line is just the slope of this line here which is one. It's rumored that the only mathematician who ever managed to answer Arnold's question was Fulton's who answered almost immediately. And when someone asked him how Fulton's done it, he said he guessed from the way Arnold was talking that the answer was something really simple. So he just guessed the simplest number he could think of which was one. So the point of this example is that sometimes thinking about a calculus problem geometrically is much better than trying to attack it algebraically.