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? 

$\frac{1}{2}$ 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 $\frac{1}{2}$.
