5. Supremums. At the start of this section, we determined that the sequence of numbers 0.9, 0.99, 0.999, 0.9999,... approaches 1 as a limit, since it can be argued that each number in the sequence is \"closer\" to 1 than the previous number in the sequence. A supremum is any number that is greater than all numbers of a sequence. Thus, 1 is a supremum of this sequence. In fact, any number $S > 1$ is a supre- mum of this sequence.
a) Consider the sequence $1 - 0.9, 1 - 0.99, 1 - 0.999,...$ What is the limit of this sequence?
b) Now consider the sequence $3 - 0.9, 3 - 0.99, 3 - 0.999,...$ What is the limit of this sequence?
c) Based on parts (a) and (b), what extra requirement should be placed on a supremum $S$ in order for it to be the limit of a sequence?