I've tried to solve this limit without success. Any good idea?
$$\lim_{x \to \infty} (x-1)\arctan (x)-\frac{\pi}{2}x$$
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what is $\pi(x)$? – Math-fun Jun 22 '17 at 14:46
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What is the function $\pi(x)$ ? – Evargalo Jun 22 '17 at 14:46
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Is that $\pi(x)$ as in the prime-counting function? – Jun 22 '17 at 14:46
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Ah, it was just $\dfrac\pi2x$. – Jun 22 '17 at 14:49
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2Writing error. Sorry. I just updated my question. – Jun 22 '17 at 14:49
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Related : https://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function – lab bhattacharjee Jun 22 '17 at 15:36
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Note that $\lim_{x\to\infty}\arctan x=\frac{\pi}{2}$, so what you need to compute is $$ \lim_{x\to\infty}\Bigl(x\arctan x-\frac{\pi}{2}x\Bigr) $$ Note that, for $x>0$, $$ \arctan x+\arctan\frac{1}{x}=\frac{\pi}{2} $$ Use this identity and substitute $t=1/x$.
egreg
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As noted by egreg, $\lim_{x\to\infty} \arctan(x)=\frac \pi 2$.
So, if you know L'Hopital's rule, you can rewrite your limit as
$$\lim_{x\to\infty}\left(\frac{\arctan(x)-\frac{\pi}{2}}{\frac1x}\right)-\lim_{x\to\infty}\arctan(x)]$$
and apply L'Hopital's rule to the first limit.
paw88789
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