(I don't believe this is classified as a 'homework question' since the solution is provided. Plus it's not really my homework. Apologies if I'm mistaken.) This question involves the asymptotic behaviour of [itex]x = \epsilon \log(1/x)[/itex] as x tends to 0. The lecturer provided a solution using the method of iteration. Basically, Afterwards, it is shown that the iterative provides the leading order behaviour of [tex]x \sim \epsilon\log(1/\epsilon)[/tex] I found this to be deeply disturbing. The truth is that x is NOT "roughly" around x = epsilon. It's a whole factor of log(1/epsilon) away. I suppose this deals with the semantics of how you're using the word "roughly", but the fact is that your initial guess for the iterative method is not the correct leading order solution! In fact, as epsilon tends to 0, your first guess is INFINITELY bad! Putting semantics aside, I'm trying to teach someone this question and motivate WHY we should be putting x = epsilon into the iterative method. I can't find a reason.