# Thread: how to prove that this function is bounded..

1. ## how to prove that this function is bounded..

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
??

2. Originally Posted by transgalactic
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 $\displaystyle f(x)<.7$

3. Originally Posted by transgalactic
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
??
As the denominator has no zeros $\displaystyle \mathbb{R}$ and the numerator is a polynomial, it is sufficient to show that the two limits as $\displaystyle x \to \pm \infty$ are both finite.

CB

4. solved it thanks