0

Is the problem $\sin(x) = O(1-x)$ as $x\to 1$ true or false?

I think that it is false, but I don't know how to prove it. I know that for some ε there exists a constant C so that $|x-a|\leq \epsilon$ implies that $|f(x)-a|\leq C|g(x)|.$

I don't know what $\epsilon$ or $C$ to choose to show that the above statement is false.

Thomas Andrews
  • 177,126
eilema
  • 11
  • You need to show that for any $\epsilon>0$ and any $C>0$ there is some $x\in(1-\epsilon,1+\epsilon$ where the inequality fails. – Thomas Andrews Sep 03 '21 at 00:51
  • 4
    Hint: $\sin(x)$ does not go to $0$ as $x$ goes to $1$. – paul garrett Sep 03 '21 at 00:51
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 03 '21 at 00:58
  • @paulgarrett I wonder if $f(n)\in\mathcal O(0)$ is also $\in\mathcal O(\sin(1))$ since technically whatever is bounded from above by $0$ is also bounded from above by $\sin(1)$. – an4s Sep 03 '21 at 01:08
  • @an4s, yes, and/but my reading of the question is that it's somewhat opposite of your a-fortiori true fact... Hard to guess what was really intended in the question. – paul garrett Sep 03 '21 at 01:14
  • One could write $\sin(x)=sin(1)+O(1-x)$, this just means that the sine is a little better than continuous in $x=1$. – Lutz Lehmann Sep 03 '21 at 04:55

1 Answers1

1

Firstly let me note that definition of $g(x) \in O(f(x)), x\to a$ is different from that you wrote: $$O(f(x)), x\to a = \{ \phi \colon \exists C>0, \exists \varepsilon>0,\forall x \ (0<|x-a|< \varepsilon \Rightarrow |\phi(x)| \leqslant C\cdot |f(x)|)\}$$

So, we need to have $|f(x)|\leqslant C|g(x)|$, not $|f(x)-a|\leqslant C|g(x)|$.

Now let's consider inequality $$|\sin x| \leqslant C|1-x|$$ and let $x \to 1$. Then right side tends to $0$, while left side gives $|\sin 1|>0$. Obtained gives $\sin(x) \notin O(1-x), x \to 1$.

zkutch
  • 13,410