Taylor/Maclaurin proof

**Stonehambey** · July 30th 2009, 09:30 AM

Hi

I would like to prove the following inequalities,

1) $f(x) = \frac{7-3\cos x-6\sin x}{9-3\cos x-8\sin x}$ , show that $\frac{1}{2}\leq f(x)\leq 1$

2) Given that $x > 0$ , prove that $x> \sin x > x - \frac{1}{6}x^3$

It's in part of a book about power series expansion of functions, so I'm pretty sure they're looking for a solution involving those. I've put in the expansions for sin and cos but I don't see the logic in how to prove these statements for all x.

A nudge in the right direction would be much appreciated, thanks

Stonehambey

**Random Variable** · July 30th 2009, 09:58 AM

$\sin x = x + O(x^{3})$
$\sin x = x -\frac{x^{3}}{3!} + O(x^{5})$
$\sin x = x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} + O(x^{7})$

So if $x>0$ , then $x$ overestimates the value of $\sin x$ and $x- \frac{x^{3}}{3!}$ underestimates the value of $\sin x$

But that's not much of a proof.

**Stonehambey** · July 30th 2009, 10:26 AM

Yeah I was thinking along the same lines, it was proving that the remaining terms do not affect it which was the bit where I was getting stuck, do I have to prove $\lim_{x \to +\infty}\frac{x^n}{n!}=0$ anywhere?

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