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Is $\sin\frac{1}{x} \lt \frac{1}{x},\ \forall x\geq 1$?

I tried "copying" the proof of $\ln x \lt x, \forall x\geq 1$ but it didn't quite work.

Here's what I did: Let $f(x)=\sin\frac{1}{x}-\frac{1}{x}$, I'd like to show that $f(x)\lt 0$. $f'(x)=\frac{1}{x^2}(1-\cos\frac{1}{x}) \gt 0$. So $f(x)$ is increasing, then $f(x)\gt f(1), \forall x \gt 1$ but $f(1)=\sin 1 - 1\lt 0$. And now I'm stuck, not sure if what I did is in the right direction.

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Can't you reformulate this as " Is $\sin x < x, \forall 0<x\leq 1$ and then apply your trick? –  Myself Jan 11 '11 at 19:49
...or look at the usual right triangle with angle $\phi$ (between 0 and $\pi/2$) in the unit circle; the side with length $\sin \phi$ is clearly shorter than the subtended arc (which has length $\phi$). (It's shorter, since a straight line is the shortest path between two points.) –  Hans Lundmark Jan 11 '11 at 20:30
What does the upside-down A mean? –  Ovi Apr 7 '13 at 14:13

3 Answers 3

up vote 4 down vote accepted

The function is increasing as you note. Now consider $$\lim_{x\to\infty} f(x) = \lim_{x\to\infty}\left(\sin\frac{1}{x} - \frac{1}{x}\right).$$ The limit is $0$; since $f$ is strictly increasing on $[1,\infty)$, has negative value at $1$, and approaches $0$ at $\infty$, can it ever be equal to zero or positive on $[1,\infty)$?

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I.e. it has a horizontal asymptote. –  PEV Jan 11 '11 at 19:55
@Trevor, I think you mean I.e, not e.g. :-) –  Asaf Karagila Jan 11 '11 at 19:55
@Trevor: "E.g." means "exempli gratia", literallty "for the sake of an example", or more informally, "for example". An example of what? –  Arturo Magidin Jan 11 '11 at 19:55
@Arturo: Thanks I always used those two interchangeably. –  PEV Jan 11 '11 at 19:58
@Trevor: Dear me, no! "e.g." is used to mean "for example"; "i.e." stands for "id est", literally "that is", and it means exactly that. As long as we are at it, "cf." is an abbreviation for "confer", meaning "compare"; it is often incorrectly used to mean "see" or "see also". –  Arturo Magidin Jan 11 '11 at 20:02

$f'(x)=0$ precisely when $\cos\left(\frac{1}{x}\right)=1$, which happens when $\frac{1}{x}\in \{2\pi n\,\vert\,n\in \mathbb{N}\}$ (since you want $x\geq 1$). So, you have infinitely many critical points ($\frac{1}{2\pi},\pm\frac{1}{4\pi},\cdots$), none of which are in the interval $[1,\infty)$. So, for all $x\geq 1$, $f(x)\geq f(1)=\sin(1)-1<0$. Also, it's not too tough to show that $\lim\limits_{x\rightarrow \infty} \sin\left(\frac{1}{x}\right)-\frac{1}{x}=0$ (just use the fact that $-1\leq\sin\left(\frac{1}{x}\right)\leq 1$). So, we have a differentiable function that's negative at $x=1$, strictly increasing, and has a limit as $x\rightarrow \infty$ of zero. Thus, it's negative on $[1,\infty)$.

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How are you using the fact that $\sin$ is bounded to conclude that $\lim\sin\frac{1}{x} - \frac{1}{x}=0$ as $x\to\infty$? It follows from continuity and the value of $\lim\frac{1}{x}$, not from boundedness. –  Arturo Magidin Jan 11 '11 at 20:08

I think the statement ($\sin\frac{1}{x} \lt \frac{1}{x},\ \forall x\geq 1$) is equivalent to saying $$\sin x \lt x,\ \forall x \text{ such that } 0<x\leq 1$$

And well, letting $f(x)=x-\sin x$ we've got $f(0)=0$. Then $f'(x) = 1-\cos{x}$, so that $f'(x)>0$ for all $x$ such that $0<x\leq 1$, so your claim is true.

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