Hello. I hope this is the right place to post this:
why is ⎮x⎮÷ (1+x^2)
bounded at 1/2 , meaning ≤ 1/2?
thanks a lot
notice that:
$\displaystyle
\mathop {\lim }\limits_{\left| x \right| \to \infty } \frac{{\left| x \right|}}
{{1 + x^2 }} = 0$
you can also easily find that the function has two symmetrical maxima points at 1 and -1
f(1) = f(-1) = 0.5 from the combination of these two facts we conclude that:
$\displaystyle
\frac{{\left| x \right|}}
{{1 + x^2 }} \leqslant \frac{1}
{2}\quad \forall x \in \Re
$