F(x)=cos(x)/x
Can someone help me to prove that this function converges to +infinity when x->0^{+ (With epsilon and delta)}
You are misusing the term "converges". A function that tends to infinity diverges. I believe you are trying to show the limit exists as an infinity limit.
You should know some properties of . For instance, you should know that is a strictly decreasing function on the interval . Since , you know that for all . Hence, for all . Now, given any , find such that implies .
Intuitively, what is happening is this:
cos(x)/x = (cos(x))(1/x).
The first factor is going to be bounded (by 1 above, and some positive number > 0 below) on an interval (0,δ) if δ is small enough. The second factor is going to get very large if δ is small enough.
So suppose we need to pick δ so that cos(x)/x > M, for some (arbitrarily large) integer M.
Pick δ so that (at least), 1/δ > 2M (what does this mean? It means that δ < 1/(2M), of course).
Now we need to perhaps further restrict δ so that cos(δ) > 1/2. As SlipEternal showed, this means choosing δ < pi/3 (there is nothing special about pi/3, it's just that we KNOW the value of cos(pi/3), and pi/3 lies in the interval (0,pi/2)...we could use ANY number in this interval, but coming up with a value for cos at this number might not be so much fun. pi/6 or pi/4 would work just as well, but the math would be "messier").
So if δ < min(1/(2M),pi/3), we have:
cos(x)/x > cos(x)/δ > cos(δ)/δ > (1/2)/(1/(2M)) = 1/(1/M) = M.
(I state it this way, using M instead of ε, because ε is usually used for a very small number, and we want to show that cos(x)/x near 0 (but on the positive side) is very LARGE, in fact, if we get "close enough" we can make it larger than any positive integer M, so it must be infinite).