Finding the limit of: $\lim_{x\rightarrow +\infty}\left(x\arctan x-x\dfrac{\pi}{2}\right)$ $\lim_{x\rightarrow +\infty}\left(x\arctan x-x\dfrac{\pi}{2}\right)$
I just removed a lot of unnecessary text from this post. If anyone could tell me how to find this limit, without L'Hôpital's rule, that would be awesome.
 A: $$\frac{\pi}{2}=\arctan(x)+\arctan\left(\frac{1}{x}\right)$$
then
$$x\arctan x-x\frac{\pi}{2}=-x\arctan\left(\frac{1}{x}\right)=-\frac{\arctan(\frac{1}{x})}{\frac{1}{x}}$$
Then,
$$\lim_{x\to\infty }-\frac{\arctan(\frac{1}{x})}{\frac{1}{x}}=\lim_{u\to 0}-\frac{\arctan u}{u},$$
First answer: using series.
$$\arctan(x)= x+x\varepsilon(x)\quad\text{if }x\to 0$$
where $\lim_{x\to 0}\varepsilon(x)=0$,
then,
$$\lim_{u\to 0}-\frac{\arctan u}{u}=-\lim_{u\to 0}\frac{u+u\varepsilon(u)}{u}=-1-\lim_{u\to 0}\varepsilon(u)=-1.$$
Second answer: using definition of derivate.
$$\lim_{u\to 0}-\frac{\arctan u}{u}=-\lim_{u\to 0}\frac{\arctan(u)-\arctan(0)}{u-0}=-\arctan'(0)=-1.$$
A: Using idm wonderful answer, we get that $\displaystyle \lim_{x\to\infty}\left(x \arctan(x)-x \frac{\pi}{2}\right)=\lim_{u\to0}-\frac{\arctan(u)}{u}$.
Let $\arctan(u)=v$, thus $u=\tan(v)$ and $v\xrightarrow{u\to0}0$.
We get $$\small\lim_{u\to0}-\frac{\arctan(u)}{u}=\lim_{v\to0}-\frac{v}{\tan(v)}=\lim_{v\to0}-\frac{v \cos(v)}{\sin(v)}=\lim_{v\to0}-\frac{\cos(v)}{1}\cdot{\lim_{v\to0}\frac{v}{\sin(v)}}=-1\cdot{1}=-1$$
A: Draw a triangle representing the relation $u = \arctan{x}$; that is, one with the leg adjacent to $u$ being 1, and the opposite leg being $x$. Then $\tan{u}=x$, and
we can write the limit you're interested in as
$$
\lim_{u \to \pi/2} ((u-\frac{\pi}{2})\tan{u})
$$
Writing the tangent as a ratio of sine and cosine and making the substitution $v=u-\pi/2$, we obtain
$$
\lim_{v\to 0} \left (\frac{\sin(v+\pi/2)}{\cos{(v+\pi/2)}}v\right) = \lim_{v\to 0}\left( \frac{\cos{v}}{-\sin{v}} v \right) = \lim_{v\to 0} \left( -\cos{v} \cdot \frac{v}{\sin{v}}\right) = -1\cdot 1 = -1
$$
where we used the cosine and sine addition formulas (or even just by investigating their relative phases) to get the second equality, and the fact that $\lim_{t\to 0} \frac{t}{\sin{t}} =1$ for the last step.
