# Evaluate $\lim\limits_{x\to\infty}x(\frac{\pi}{2}-\arctan(x))$ without using L'Hôpital

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

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

related :

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

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

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

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

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

Notice that $$\frac{\arctan y }{y} = \frac{\arctan y}{\tan (\arctan y)} = \frac{\arctan y}{\sin (\arctan y)} \cos(\arctan y) \to 1$$ as $y\to 0$.
• why $\frac{\arctan(1/x)}{1/x}\to 1$ ?
• Do you know that $\sin(t) / t \to 1$ as $t\to 0$? Dec 20 '15 at 12:34
• @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 $0$. Dec 20 '15 at 12:54
$$\lim_{x\rightarrow \infty }x\left ( \frac{\pi}{2}-\operatorname{arctg}x \right )=\lim_{x\rightarrow \infty }\frac{\frac{\pi}{2}-\operatorname{arctg}x}{1/x}=\lim_{x\rightarrow \infty }\frac{x^2}{x^2+1}=\lim_{x\rightarrow \infty }\frac{1}{1+\frac{1}{x^2}}=1$$