f(x)=(1+x+x^2 +x^3 +x^4)/(1+x^4)
using lmit n->+infinity
obviously this function goes to 1
is that all i need to do in order to show that its bounded
??
Case #1 $\displaystyle 1\leqslant x$
$\displaystyle \begin{aligned}\frac{1+x+x^2+x^3+x^4}{1+x^4}&\leqs lant\frac{x^4+x^4+x^4+x^4+x^4}{1+x^4}\\
&\leqslant\frac{5x^4}{x^4}\\
&=5\end{aligned}$
Case #2 $\displaystyle 0\leqslant x <1$
$\displaystyle \begin{aligned}\frac{1+x+x^2+x^3+x^4}{1+x^4}&\leqs lant\frac{1+x+x+x+x}{1+x^4}\\
&\leqslant\frac{4+4x}{1+x^4}\\
&\leqslant\frac{4(1+x)}{1-x^4}\\
&=\frac{4}{1-x+x^2-x^3}\\
&\leqslant 4\end{aligned}$
Is that what you wanted?
EDIT: Using calculus you can greatly improve this to $\displaystyle f(x)<.7$