# Runtime analysis: lower limit of binary search in the worst case

For the next question, it is requested to prove that the lower limit in the worst case is log (n). I have no problem in trying this and the solution makes 100% sense to me. However, there is a comment at the bottom of the solution that asks why we had to assume that if the loop does not return before, then it runs for the omega log (n) iterations. Why is not it obvious based on the fact that the upper limit in the worst case is log (n)? I hope someone can help.