Q:

Stability and complexity of some functions

Can someone check if my solutions/arguments on this exercise are correct?  Thanks! 

Are the following statements true or false? 

$\sin (x)=\mathcal{O}(1)$ as $x \rightarrow \infty$
$\sin (x)=\mathcal{O}(1)$ as $x \rightarrow 0$
$n!=\mathcal{O}((\frac{n}{e})^n)$ as $x \rightarrow \infty$

I think that the first one is false, because if we approximate the $sin$ function using a truncated taylor series and we substract the real $sin$ function of this, it could still get arbitrarily large. 
The second one will be true, i think, by using the same argument as above.  Because if we take the limit to $0$, the order will be constant. 
For the last one I have no idea. 

A:

Contrary to your thoughts, part 1 is true: Your argument with the Taylor series does not apply.  
Simply use the definition: $\sin (x)=\mathcal{O}(1)$ as $x \rightarrow \infty$,
if there is a positive constant $M$ with $|\sin(x)| <= M|1|$ for all $x$ greater than some $x_0$.  In this case you can take $M=1, x_0=0$. 
Part 2 uses the same argumentation. 
For part 3: Here you can use Stirling's approximation of the factorial $n!$
which says
$$1 \le \frac{n!}{\sqrt{2\pi n}(\frac{n}{e})^n} \le \frac{e}{\sqrt{2\pi}}\cdot$$ And this shows that $n!=\mathcal{O}(\sqrt{n}(\frac{n}{e})^n)$ as $n\rightarrow \infty$ and not $n!=\mathcal{O}((\frac{n}{e})^n)$, i.e. part 3 is false.