Find $\sup f^{n}(x)=\sup \underbrace{f\circ f\circ\cdots\circ f}_{n \text{ times}}(x)$. Let $f(x)={x\over \sqrt{1+x^2}}$. Find $\sup f^{n}(x)=\sup \underbrace{f\circ f\circ\cdots\circ f}_{n \text{ times}}(x)$. $n\in \Bbb{N}.$
I suppose $\sup f=1$. Isn't $\sup f^{n}(x)=1$ directly? I am not so sure what to do here...I would appreciate your help.
This question is not to be answered using derivatives(i.e it was given to me before I studied about that), limits or continuity. What it relies on so far is sup\lim and boundness...
 A: Observe that $f$ is increasing as you can see by computing its derivative. Therefore $f^n$ is also increasing. Hence $\sup_x f^n(x)=\lim\limits_{x\to+\infty}f^n(x)$.
Since $f$ is continuous you can start computing that limit from inside out.
$\lim\limits_{x\to+\infty} f(x)=1$. Then $\lim\limits_{x\to+\infty} f^2(x)=f(1)=\frac{1}{\sqrt{2}}$, $\lim\limits_{x\to+\infty}f^3(x)=f\left(\frac{1}{\sqrt{2}}\right)$ ... etc.

You don't need derivatives to show the function is increasing. We just need to compare, for $x<y$ the values of $f$. With a small change to the expression we see that $$f(x)=\frac{\text{sgn}(x)}{\sqrt{1+\frac{1}{x^2}}}<\frac{\text{sgn}(y)}{\sqrt{1+\frac{1}{y^2}}}=f(y).$$ The inequality follows since each operation involved to compute the expression is increasing ($x^2$, $1/x$, $1+x$, $\sqrt{x}$, $\text{sgn}(x)$).
In the same way, you don't need limits or continuity explicitly. But they will be there anyway in disguise. Since $\frac{1}{x^2}$ is decreasing and takes values arbitrarily close to zero, then $f(x)=\frac{\text{sgn}(x)}{\sqrt{1+\frac{1}{x^2}}}$ takes values arbitrarily close to $1$. Since $f(x)\leq1$ that is its supremum. Now continue composing this with $f$'s.

A: Hint: use induction on $n$ to prove that $f^n(x)=\frac{x}{\sqrt{1+nx^2}}$.
