Question 6 section 8.1 of Introduction to real analysis by Bartle and Sherbert.

Show that $lim(Arctan nx) = (pi/2)sgn x$ for $x$ in R, $x>=0$.

I have a final coming up and I've started doing some of the exercises in the book. I'm not sure how to go about proving this. I know I'm trying to show |Arctan(nx)-$(pi/2)sgn| < \epsilon$ but the sgn function is throwing me off. Any help is appreciated.


Hint: Show the problem reduces to determining $\lim_{x \rightarrow \infty} \arctan(x)$ and $\lim_{x \rightarrow -\infty} \arctan(x)$.

  • $\begingroup$ How's this? If x > 0, for any epsilon > 0, We see that for any n > (tan(pi/2 - epsilon))/x, we have pi/2 > arctan(nx) > arctan( tan(pi/2-epsilon)) where it follows that |arctan(nx)-pi/2| < epsilon. so lim(arctan(nx))=pi/2 for x > 0 and for x<0 -x >0 so -lim(arctann(-x))= -pi/2 $\endgroup$ – User38 Apr 20 '14 at 21:16
  • 1
    $\begingroup$ Yes that works. $\endgroup$ – Zarrax Apr 20 '14 at 21:29

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