how to prove that this function is bounded..

**transgalactic** · December 14th 2008, 01:52 PM

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
??

**Mathstud28** · December 14th 2008, 02:23 PM

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 $1\leqslant x$

$\begin{aligned}\frac{1+x+x^2+x^3+x^4}{1+x^4}&\leqs lant\frac{x^4+x^4+x^4+x^4+x^4}{1+x^4}\\ &\leqslant\frac{5x^4}{x^4}\\ &=5\end{aligned}$

Case #2 $0\leqslant x <1$

$\begin{aligned}\frac{1+x+x^2+x^3+x^4}{1+x^4}&\leqs lant\frac{1+x+x+x+x}{1+x^4}\\ &\leqslant\frac{4+4x}{1+x^4}\\ &\leqslant\frac{4(1+x)}{1-x^4}\\ &=\frac{4}{1-x+x^2-x^3}\\ &\leqslant 4\end{aligned}$

Is that what you wanted?

EDIT: Using calculus you can greatly improve this to $f(x)<.7$

**CaptainBlack** · December 14th 2008, 02:30 PM

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 $\mathbb{R}$ and the numerator is a polynomial, it is sufficient to show that the two limits as $x \to \pm \infty$ are both finite.

CB

**transgalactic** · December 21st 2008, 02:23 AM

solved it thanks

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