3
$\begingroup$

This is part of a larger problem, but now I'm trying to find $$\lim_{n\to\infty}\left(\frac12 + \frac1{\pi}\arctan\left(\frac{nx}{t}\right)\right)^n.$$ I think I have to use L'Hôpital's, but I'm not sure how.

  • 0
    Yes, you can just type standard LaTeX, enclosed in `$` for in-line formulas and in `$$` for displayed formulas.2011-04-27
  • 0
    You don't say the limit as what goes where, but I assume it's the limit as $n\to\infty$.2011-04-27
  • 0
    Limit as what is going where?2011-04-27
  • 0
    yes, limit as n goes to infinity2011-04-27
  • 0
    Is it perhaps $$ (1/2 - (1/\pi) arctan(nx/t))^n $$ instead?2011-04-27
  • 0
    Hint: Let $u_n=\left(\frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{nx}{t}\right)\right)-1$. Since $\lim_{n \to \infty} u_n=0$, then $\lim_{n \to \infty} (1+u_n)^\frac{1}{u_n}= e$. Then use L'H..... You can also threat it as a general $1^\infty$ L'H limit.2011-04-27
  • 0
    Do you mean use L'H on $u_n$?2011-04-27
  • 0
    I got something like $\exp(\pm t/x\pi)$ by rewriting the arctangent as the arc tangent of something that goes to zero, and then using the series for arctangent around zero.2011-04-27
  • 0
    @mixedmath: Instead of what?2011-04-27
  • 0
    @Carl: I got the same thing using rewriting and L'Hopital's (with the minus sign).2011-04-27
  • 0
    @Arturo: Aha, I had made a silly mistake. I placed a negative sign in front because I had thought (mistakenly) that the limit clearly went to infinity. I now agree with you and Carl.2011-04-27
  • 0
    @Carl, that's the correct answer. When you say you rewrote the arctangent, how did you rewrite it into something that goes to zero?2011-04-27
  • 0
    nevermind, got it2011-04-27

2 Answers 2