Is $\sin\frac{1}{x} \lt \frac{1}{x},\ \forall x\geq 1$? 
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.
 A: 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 $[1,\infty)$?
A: $f'(x)=0$ 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 $[1,\infty)$.  
A: 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. 
