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 November 18th, 2018, 08:28 PM #1 Senior Member   Joined: Apr 2017 From: New York Posts: 162 Thanks: 6 limit Hi everyone, how can we take the limit of this function. I don't need the asnwer. I have it. I just need to know the steps. lim(x is app to infinity) (-1)^(x+1)*x/ x+(sqrtx) Last edited by Leonardox; November 18th, 2018 at 08:36 PM.
 November 18th, 2018, 09:59 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,584 Thanks: 1430 $\lim \limits_{x \to \infty}~(-1)^{x+1}\cdot \dfrac{x}{x+\sqrt{x}}$ I don't believe this limit exists. Divide top and bottom by $x$ $\lim \limits_{x \to \infty}~(-1)^{x+1}\cdot \dfrac{1}{1+\frac{1}{\sqrt{x}}}$ clearly $\lim \limits_{x\to\infty}~\dfrac{1}{\sqrt{x}} = 0$ so as $x \to \infty$ you start ping ponging back and forth between terms very close to 1 and terms very close to -1. This never converges, the two terms just get closer and closer to 1 and -1. Thanks from topsquark
 November 18th, 2018, 11:34 PM #3 Senior Member   Joined: Oct 2009 Posts: 884 Thanks: 340 Also, if you write $x$, then to me that is a continuous variable, meaning all $x\in \mathbb{R}$ are allowed. In that case though, isn't $(-1)^{x+1}$ ill-defined? Thanks from topsquark Last edited by greg1313; November 20th, 2018 at 01:28 AM.
 November 19th, 2018, 04:18 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 728 Thanks: 98 For x-integer then (-1)^x is defined
 November 20th, 2018, 05:58 PM #5 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 $(-1)^x=e^{x\pi i}$

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