calculus - How to solve the limit $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$

Next week I have a math exam. While I was doing some exercises I came across this interesting limit:

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

After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:

$\lim\limits_{x\to \infty} (x \arctan x - \frac{x\pi}{2}) = \lim\limits_{x\to \infty} \frac{2x^2\arctan x - x^2\pi}{2x} \stackrel{(H)}{=} \lim\limits_{x\to \infty} \frac{4x\arctan x - \frac{2}{x^2+1}-2\pi x+2}{2} = \lim\limits_{x\to \infty} \frac{4x^2\arctan x - \frac{2x}{x^2+1}-2\pi x^2+2x}{2x} \stackrel{(H)}{=} \lim\limits_{x\to \infty} \frac{8x\arctan x - \frac{2x^2+6}{(x^2+1)^2}-4\pi x+6}{2} = \dots$

This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?

Answer

Observe

$$\begin{align} \lim_{x\rightarrow \infty} x\arctan x-x\frac{\pi}{2}=\lim_{x\rightarrow\infty}\frac{\arctan x-\frac{\pi}{2}}{x^{-1}} = \lim_{x\rightarrow \infty} \frac{\frac{1}{1+x^2}}{-x^{-2}}=-1 \end{align}$$