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Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1Are you allowed to use any series?

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1then I recommend using the series for tan^(1)x, because I know that this series approaches for very large x the value pi / 2, so you can just take a multiply (2 times) of that.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1but that's only my first intuition and maybe it doesn't answer that problem rigorously

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \lim_{x \to \infty} \tan^{1} x = \frac{\pi}{2} \] \[ \lim_{x \to \infty} 2 \tan^{1}x = 2 \sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}x^{2n+1}\]

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \large \lim_{x \to \infty} 2 \tan^{1}x =\lim_{x \to \infty}\ 2 \sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}x^{2n+1} \] Shouldn't forget to carry out the limit

Neemo
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0}^{+ \infty }\frac{(1)^n}{2n+1}=\frac{ \pi}{4}\]
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