asymptotically sharp upper and lower bound for for arctan [closed]
Hint: $$\dfrac{1}{x^2} \ge \dfrac{d}{dx} \arctan(x) = \dfrac{1}{1+x^2} \ge \ldots$$
Hint:
Consider $f(x) = \tan^{-1}x + \frac1x-\frac\pi 2$, show $f'(x)< 0$ and note $\lim_{x\to \infty} f(x) = 0$.
Next consider $g(x) = \frac\pi2 - \frac1x-\tan^{-1}x +\frac1{3x^3}$ along the same lines.
We have $$ \arctan(1/x)=\frac\pi2-\int_0^x\frac{\mathrm{d}t}{1+t^2} $$ and $$ 1-t^2\le\frac1{1+t^2}\le1 $$ Therefore, $$ \frac\pi2-x\le\arctan(1/x)\le\frac\pi2-x+\frac13x^3 $$ and $$ \frac\pi2-\frac1x\le\arctan(x)\le\frac\pi2-\frac1x+\frac1{3x^3} $$