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Let f(x) = $\frac{x}{|x|}$ if x $\neq$ 0, and define f(0)=0. Show that f(x) = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)$


My work:
$\frac{x}{|x|}$ = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)$
$\frac{x}{|x|}$ = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \frac{cos(nx)}{sin(nx)}$
$\lim_{n \rightarrow \infty} cos(n)$ = -1 and 1, likewise $\lim_{n \rightarrow \infty} sin(n)$ = -1 and 1.
So when $\frac{-1}{-1}$ and $\frac{1}{1}$ then the limit is equal to 1. Likewise when $\frac{-1}{1}$ and $\frac{1}{-1}$ then the limit is equal to -1.
f(x) always equals either 1 or -1 so they can be confirmed to be equivalent. However, I'm having problems with the $\frac{2}{\pi}$ and the effects of x in the limit.

Hints please! Thanks!

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    $\begingroup$ Note $\tan^{-1}(nx)$ does not equal $\cos(nx)/sin(nx)$. $\endgroup$ – kobe Jan 20 '15 at 23:39
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    $\begingroup$ The limits $\lim \limits_{n\to \infty}\left(\cos(n)\right)$ and $\lim \limits_{n\to \infty}\left(\sin(n)\right)$ do not exist, contrarily to what you say. Do you know what $\lim \limits_{t\to +\infty}\left(\tan^{-1}(t)\right)$ is? $\endgroup$ – Git Gud Jan 20 '15 at 23:40
  • $\begingroup$ To both, I guess I don't understand what n means because I just assumed it was a variable similar to x meaning I think the limit of cos(x) is bouncing between -1 and 1 though I understand that an actual limit should be one defined number that it approaches. Then also, by that I don't understand why tan^{-1}(nx) does not equal the cos/sin because I took tan A, where A = nx to come to that conclusion. Either way it still leaves me stuck. $\endgroup$ – Bee Jan 20 '15 at 23:44
  • $\begingroup$ To Git Gud, How can n approaches infinity exist in a limit where n is not a variable? (Also, I assume likewise it has no limit since it bounces between -$\infty$ and $\infty$) $\endgroup$ – Bee Jan 20 '15 at 23:46
  • $\begingroup$ One has $\lim \limits_{t\to +\infty}\left(\tan^{-1}(t)\right)=\dfrac \pi 2$. How does one go about seeing this? That depends on the exact definitions of the trigonometric functions. But I think (and I'm just guessing here) that you're supposed to solve is graphically. You should know what the graph of $\tan$ looks like and you need to investigate how to get the graph of $\tan^{-1}$ from the graph of $\tan$. After getting the graph, the limit will present itself. $\endgroup$ – Git Gud Jan 20 '15 at 23:51
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If $x = 0$, $\frac{2}{\pi} \tan^{-1}(nx) = 0$ for all $n$, and thus $\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = 0$ when $x = 0$. If $x > 0$, then $$\lim_{n\to \infty} \frac{2}{\pi} \tan^{-1}(nx) = \frac{2}{\pi}\lim_{u\to +\infty} \tan^{-1}(u) = \frac{2}{\pi}\cdot\frac{\pi}{2} = 1 = \frac{x}{|x|}.$$ If $x < 0$, then $$\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = \frac{2}{\pi} \lim_{u\to -\infty} \tan^{-1}(u) = -1 = \frac{x}{|x|}.$$ Hence, for all $x\neq 0$, $\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = \frac{x}{|x|}$.

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Here is an approach.

Recall that $$ \tan^{-1} x +\tan^{-1} \frac1x = {\rm{sign}}\:x \: \cdot\frac \pi 2, \qquad x\neq 0, \tag1 $$ and that, for $u$ near $0$, $$ \tan^{-1} u = u +\mathcal{O}\left(u^3\right). \tag2 $$ Then, using $(1)$ and $(2)$, we have, for $x \neq 0$, $$ \begin{align} \lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)&= \lim_{n \rightarrow \infty} \frac{2}{\pi} \left({\rm{sign}}\:nx \: \cdot\frac \pi 2- \tan^{-1}\left(\frac{1}{nx}\right)\right)\\\\ &={\rm{sign}}\:x - \frac{2}{\pi}\lim_{n \rightarrow \infty} \tan^{-1}\left(\frac{1}{nx}\right)\\\\ &={\rm{sign}}\:x - \frac{2}{\pi}\lim_{n \rightarrow \infty} \left(\frac{1}{nx}\right) \\\\ &={\rm{sign}}\:x \\\\ &=\frac{x}{|x|}. \end{align} $$

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