Prove that $\frac{1}{x} = O\left(\frac{\pi}{2} - \arctan(x)\right)$ as $x\to\infty$.
I am confused how we used the formal definition of big-$O$-notation to prove this. Notice that this is just $\frac{1}{x}$.
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2Hint $\frac{\pi}{2} - \arctan(x) = \arctan(1/x)$ when $x>0$. – Gribouillis Jan 16 '21 at 22:03
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1Maybe it helps $\lim\limits_{t\to 0}\frac{\tan(t)}t=1$ where $t=\arctan\left(\frac1x\right)$. Related post. – PinkyWay Jan 16 '21 at 22:11
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2Are you familiar with $\lim_{x\to\infty}f(x)/g(x) \neq \infty \implies f = O(g)$? Because a straightforward application of L'Hopital's rule will suffice to show that. – DMcMor Jan 16 '21 at 22:11
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For $x>1,$
$$0<\pi/2 - \arctan x = \int_x^\infty \frac{1}{1+t^2}\,dt > \int_x^\infty \frac{1}{2t^2}\,dt =\frac{1}{2x}.$$
This implies $\dfrac{1}{x} < 2(\pi/2 - \arctan x)$ and proves the result.
zhw.
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This proves $\frac\pi2-\arctan x \in O\left(\frac1x\right)$ – Maximilian Janisch Jan 16 '21 at 22:47
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@MaximilianJanisch One could use $\frac{1}{(1+t)^2}$ to get the other way. – Gribouillis Jan 16 '21 at 22:49
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1@MaximilianJanisch You're right, I was proving the reversed inequality. Now corrected. Thanks. – zhw. Jan 19 '21 at 15:40