How is $\frac{1}{2n} \leq \sin (\frac{1}{n}) \leq \frac{1}{n} $

Question

How is $\frac{1}{2n} \leq \sin (\frac{1}{n}) \leq \frac{1}{n} $

I know that $\sin \theta \leq \theta, \theta$ very small

But if take $ f(x) = \sin (\frac{1}{x}) - \frac{1}{2x}, f'(x) = \frac{-1}{x^2} \cos (\frac{1}{x}) + \frac{1}{2x^2} $

But if i take $x=\frac 4\pi, x \in (0, \pi/2), $ i am getting $f'(x) < 0$ which should be other way around. Am I missing something? I got this doubt while reading Does $\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right) $ absolutely converge?

If i say since $\sin (x)$ converges to $x$, i will have $\sin x$ values slightly less than $x$ and slightly more than $x$. But it depends on whether $f(x)$ is increasing/decreasing (local maxima or local minima)

Pls clarify

Answer 1

We know that for $x=\frac1n>0$ $$x-\frac16x^3\le \sin x \le x$$ [ refer to Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$ ]

and

$$\frac12x\le x-\frac16x^3 \iff \frac12x-\frac16 x^3\ge 0 \iff x^2\le 3 \iff0<x<\sqrt 3$$

which is true since $0<x=\frac1n \le 1$.

Answer 2

A negative value of derivative $f'$ at some point $x_0$ does not necessarily leads to negative value of the original function $f(x_0)$. It may however indicate that the function is approaching a value from above asymptotically, if the limit is left-limit. If it is a right-limit, everything is reversed.

Answer 3

For $0\le x\le\frac\pi2$, in this answer it is shown that $\tan(x)\ge x$. Therefore, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin(x)}x &=\frac{\cos(x)}{x^2}\,(x-\tan(x))\\ &\le0 \end{align} $$

Thus, $\frac{\sin(x)}x$ is decreasing on $\left[0,\frac\pi2\right]$, and since $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, $$ \frac2\pi\le\frac{\sin(x)}x\le1 $$

Answer 4

Proof

Denote $\dfrac{1}{n}=x,(n=1,2,\cdots)$. Thus, $0<x \leq 1<\dfrac{\pi}{2}$. This shows that the angle with the radian $x$ is a first-quadrant angle on the unit circle. As the figure shows, $\angle AOB=x$，which is bisected by $OD$. Thus,$$|\widehat{AEB}|=x,~~~|\widehat{AE}|=\frac{1}{2}x,~~~ |BC|=\sin x.$$

Notice that $$|\widehat{AEB}|>|AB|>|BC|>|BG|=|DA|>|\widehat{AE}|.$$ Thus, we may have $$x>\sin x>\frac{1}{2}x,$$which is desired.