November 18th, 2018, 09:28 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 119 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 09:36 PM. 
November 18th, 2018, 10:59 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,324 Thanks: 1233 
$\lim \limits_{x \to \infty}~(1)^{x+1}\cdot \dfrac{x}{x+\sqrt{x}}$ 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. 
November 19th, 2018, 12:34 AM  #3 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 247 
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 $(1)^{x+1}$ illdefined?
Last edited by greg1313; November 20th, 2018 at 02:28 AM. 
November 19th, 2018, 05:18 AM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 392 Thanks: 62 
For xinteger then (1)^x is defined

November 20th, 2018, 06:58 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,686 Thanks: 661 
$(1)^x=e^{x\pi i}$


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