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$$\lim\limits_{x\to\infty}x(\frac{\pi}{2}-\arctan(x))$$

I want to evaluate the limit without using L'Hôpitals Rule

related :

How to evaluate the following limit? $\lim\limits_{x\to\infty}x\left(\frac\pi2-\arctan x\right).$

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$\dfrac\pi2-\arctan x=y\implies x=\cot y$

As $x\to\infty,\dfrac\pi2-y\to\dfrac\pi2\iff y\to0$

$$\lim_{x\to\infty}x\left(\dfrac\pi2-\arctan x\right)=\lim_{y\to0}\dfrac y{\sin y}\cdot\lim_{y\to0}\cos y=?$$

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$$ x (\pi/2 - \arctan x) = x \arctan\frac 1 x = \frac{\arctan(1/x)}{1/x}\to 1 $$ as $x\to +\infty$.

added.

Notice that $$ \frac{\arctan y }{y} = \frac{\arctan y}{\tan (\arctan y)} = \frac{\arctan y}{\sin (\arctan y)} \cos(\arctan y) \to 1 $$ as $y\to 0$.

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  • $\begingroup$ why $\frac{\arctan(1/x)}{1/x}\to 1 $ ? $\endgroup$ – user238874 Dec 20 '15 at 11:45
  • $\begingroup$ Do you know that $\sin(t) / t \to 1$ as $t\to 0$? $\endgroup$ – Emanuele Paolini Dec 20 '15 at 12:34
  • $\begingroup$ yes, I know.... $\endgroup$ – user238874 Dec 20 '15 at 12:36
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    $\begingroup$ @Raphael: when $u$ tends to $0$, $\dfrac{\arctan u}u=\dfrac{\arctan u-\arctan 0}{u-0}$ is the variation rate at $0$, and hence tends to the value of the derivative at $0$. $\endgroup$ – Bernard Dec 20 '15 at 12:54

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