How to show that each member of a sequence of functions is bounded? The question I am working on is: Let $f_n(x) := \dfrac{nx}{1 + nx^2}$ , for $x ∈ A := [0, ∞)$. Show that each $f_n$ is bounded on $A$.
I started by taking the derivative to get $f_n'(x)$= $n(1+nx^2)-2n(nx)$ but I'm not sure whether this is the right direction. I then noticed that $0<f_n(x)<nx$, but I'm not sure if this counts as an upper bound because it still has both of the variables $n$ & $x$ in it. Any help/direction would be appreciated!
 A: I think its more clear to break into two cases.  First, consider $x \geq 1$.  Then,
$$f_n(x) = \frac{x}{1/n+x^2} < \frac{x}{x^2} = \frac{1}{x}.$$  That is, for $x \geq 1$, the function is bounded above by the function $1/x$ which is in turn bounded by $1/1 = 1$.  Now, consider $0 < x <1$.  Then, we let $y = 1/x$, so that $y >1$.  Then,
$$f_n(x) = f_n(1/y) = \frac{n}{y+n/y} < \frac{n}{y} < n.$$
Finally, for $x = 0$, the function has value of zero, so bounded.
This way you avoid derivatives, but the answer above gives you a tighter bound.
A: If $0\le x < 1$ then
$$
f_n(x)
=
\frac{nx}{1 + nx^2}
\le
nx < n
$$
and if $x\ge 1$ then
$$
f_n(x)
=
\frac{nx}{1 + nx^2}
\le
\frac{nx^2}{1 + nx^2} < 1
$$
so $f_n$ is bounded above by $n$.
A: You had a good idea with derivatives. Note that $$ f_n'(x) = \frac{n(1+nx^2) - 2nx(nx)}{(1+nx^2)^2} = \frac{n-n^2x^2}{(1+nx^2)^2}.$$
It has unique zero point (on the set $A$) at a point $x = \frac{1}{\sqrt{n}}$. Moreover, $f_n'(x) >0$ for $x \in [0,\frac{1}{\sqrt{n}})$ and $f_n'(x) < 0$ for $x \in (\frac{1}{\sqrt{n}},\infty)$, hence $x = \frac{1}{\sqrt{n}}$ is a maximum of $f_n$. Hence $$  \sup_{x \in A} f_n(x) =  f_n(\frac{1}{\sqrt{n}}) = \frac{\sqrt{n}}{1+1} = \frac{\sqrt{n}}{2}. $$ From this we see that $f_n$ is bounded on $A$ (note that $\sqrt{n}$ is a constant for any given $f_n$, hence it isn't problematic. But as we see, $\sup_n \sup_x f_n(x) = \sup_n \frac{\sqrt{n}}{2} = \infty$, so that $(f_n)$ isn't uniformly bounded on $A$).
