1. ## bounded function

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

2. notice that:

$
\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:

$
\frac{{\left| x \right|}}
{{1 + x^2 }} \leqslant \frac{1}
$

3. but then where is the 1/2 coming from?

4. Have you read my post carefully (f(1) = f(-1) = 0.5) ?

5. oh sorry. i did not see the second part when i posted it, it did not appear. it just appeared the part on the limit. i don't know how.

with the second part added is much better. now i got it
thanks!