Deducing a supremum from a given property Consider $$ S = \bigg\{ { \frac{x^2}{1+2x^2} } :x\in\mathbb{R}\bigg\}$$ 
we may guess that $\sup A=\frac{1}{2}$ 
But how does one prove this without taking limits? 
 A: If $x = 0$ then the fraction is $0$. If $x \neq 0$, then
$$
\frac{x^{2}}{1 + 2x^{2}} = \frac{1}{\frac{1}{x^{2}} + 2} \leq \frac{1}{2}.
$$
There is no $r > 2$ such that $1/r$ is instead the supremum of the set; for every $r > 2$ there is some real $x$ such that 
$$
\frac{x^{2}}{1+2x^{2}} > \frac{1}{r},
$$
say $x := 2\sqrt{\frac{1}{r-2}}$.
A: $\frac{1}{2}$ is an upper bound for $S$. In fact
$$\frac{x^2}{1+2x^2} =\frac{1}{2} \frac{2x^2}{1+2x^2}\le\frac{1}{2}\frac{1+2x^2}{1+2x^2}= \frac{1}{2} $$
for all $x \in \mathbb R$. Moreover,
$$\lim_{x \to +\infty}\frac{x^2}{1+2x^2} =\frac{1}{2} $$
and so for any $\varepsilon >0$
$$\left|\frac{x^2}{1+2x^2}-\frac{1}{2} \right|<\varepsilon $$
for $x$ large enough. In particular, for $x$ large enough, we have
$$\frac{x^2}{1+2x^2} > \frac{1}{2}-\varepsilon, $$
so the desired supremu is indeed $\frac{1}{2}$.
A: I suppose if you show that 
$$lim_{x\rightarrow \infty} f(x)=\frac{1}{2}$$
and also 
$$f(x)<\frac{1}{2} \hspace{0.5cm} \forall x$$
you are done.
The second condition can be shown by showing that the function is monotonically increasing on $[0,\infty]$ and monotonically decreasing on $[-\infty,0)$
