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$\lim_{x\rightarrow +\infty}\left(x\arctan x-x\dfrac{\pi}{2}\right)$

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.

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  • $\begingroup$ Just re-wrote the whole post as it was very misleading to what I was looking for. $\endgroup$ – PurpleManiac Nov 7 '14 at 22:16
  • $\begingroup$ So L'Hospital's Rule is forbidden. What about series? $\endgroup$ – André Nicolas Nov 7 '14 at 22:21
  • $\begingroup$ @AndréNicolas Not sure how they work to be honest. I'm trying to stick with the common limits: en.wikipedia.org/wiki/List_of_limits $\endgroup$ – PurpleManiac Nov 7 '14 at 22:22
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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.

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$$\frac{\pi}{2}=\arctan(x)+\arctan\left(\frac{1}{x}\right)$$ 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.

$$\lim_{u\to 0}-\frac{\arctan u}{u}=-\lim_{u\to 0}\frac{\arctan(u)-\arctan(0)}{u-0}=-\arctan'(0)=-1.$$

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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 $$\small\lim_{u\to0}-\frac{\arctan(u)}{u}=\lim_{v\to0}-\frac{v}{\tan(v)}=\lim_{v\to0}-\frac{v \cos(v)}{\sin(v)}=\lim_{v\to0}-\frac{\cos(v)}{1}\cdot{\lim_{v\to0}\frac{v}{\sin(v)}}=-1\cdot{1}=-1$$

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