Do you know maximum?

The set of solutions for $x^2 < 9$ is all real $x$ satisfying $-3 < x < 3$ , or $\{x | x \in \mathbb{R}, -3 < x < 3\}$ . Does this mean the maximum is 3?

No, the answer is that the maximum doesn't exist . The definition of maximum of a subset of real numbers $S$ is an element $m$ that is in the set $S$ such that for all elements $x \in S$ , we have $m \ge x$ . $m = 3$ doesn't satisfy this definition, since 3 is not in the set $\{x | x \in \mathbb{R}, -3 < x < 3\}$ . On the other hand, any smaller $m$ won't do either, since $\frac{3+m}{2} > m$ and $\frac{3+m}{2}$ is still in the set.

A related concept is supremum , also known as smallest upper bound. A supremum of a subset of real numbers is defined similarly as above, but it lacks the requirement that $m$ is in the set $S$ . More precisely, it is a real number $m$ such that for all $x \in S$ , we have $m \ge x$ ( $m$ is an upper bound), and if $m' < m$ , then there exists $x \in S$ such that $x > m'$ (there is no smaller upper bound). In this case, it is true that the supremum of the set of solutions is 3, because the supremum doesn't require it being in the set itself.

Evaluate $\displaystyle\lim_{n\rightarrow\infty}(3-\frac{1}{n})^2$ or $\displaystyle\lim_{n\rightarrow\infty}(-3+\frac{1}{n})^2$ and expand (both) to $\displaystyle\lim_{n\rightarrow\infty}(9-\frac{6}{n}+\frac{1}{n^2})$ Despite we can evaluate the limit that it tends to be 9, infinity does not have the upper bond and the fractional value will get closer to zero but not reach zero itself (i.e. any large number $n$ inserted will contain $n+1$ , which makes the fraction even smaller and close to zero), hence there is no maximum in this regard.