1. ## Prove inequality

Show that sin(x) = O(x) for all x in R.

From the defn, we must show

$\displaystyle \exists A$ s.t. $\displaystyle |\sin(x)| \le A|x|$

We know that |sin(x)| <= 1.

But now I'm stuck.

Similarly, how would I go about showing:

$\displaystyle \exists A>0,c\in \mathbb{R}$ s.t.

$\displaystyle |e^{-x}|\le A|x^{-m}|, m\in\mathbb{Z}^+,~c<x<\infty$

What's the general method for these things? (I know $\displaystyle e^x$ is always positive, so the mod signs are redundant)

(The above is derived from $\displaystyle e^{-x}=O(x^{-m})$, m positive integer.)

2. If f(x) is uniformly continuous on [1,inf) then there exist M>0 with |f(x)|<Mx

3. A little more elementary: By the mean value theorem, for any x, taking f(x)= x- sin(x), we have $\displaystyle \frac{f(x)- f(0)}{x- 0}= f'(c)$ for some c between 0 and x. But $\displaystyle f'(x)= 1- cos(x)\ge 0$. Therefore, we have $\displaystyle f(x)= x- sin(x)\ge 0$ so $\displaystyle x\ge sin(x)$.

4. Thanks -- but you've shown that sin(x) <= x, but that doesn't necessarily imply that |sin(x)|<=|x|, since

$\displaystyle a < b \not\Rightarrow |a|<|b|$, right?

(PS. I don't think I've got enough background for uniform continuity, having not studied metric spaces or almost any functional analysis)

5. You can do something similar(or symmetrical ) to what Mr. HallsofIvy did.

6. Yes, the first part of my reply was indeed directed at HallsOfIvy. I'm not convinced that it proves what was required, for the reason I mentioned.