November 18th, 2018, 08:28 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 136 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: 2,404 Thanks: 1306 
$\lim \limits_{x \to \infty}~(1)^{x+1}\cdot \dfrac{x}{x+\sqrt{x}}$ 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 18th, 2018, 11:34 PM  #3 
Senior Member Joined: Oct 2009 Posts: 771 Thanks: 276 
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 01:28 AM. 
November 19th, 2018, 04:18 AM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 479 Thanks: 73 
For xinteger then (1)^x is defined

November 20th, 2018, 05:58 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689 
$(1)^x=e^{x\pi i}$


Tags 
limit 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
upper limit = lower limit implies convergence  zylo  Calculus  13  May 31st, 2017 12:53 PM 
limit  goodfeeling  Calculus  2  October 21st, 2012 02:51 PM 
what is the limit of this? .  nappysnake  Calculus  3  November 28th, 2011 02:12 AM 
Limit Superior and Limit Inferior  veronicak5678  Real Analysis  4  August 22nd, 2011 10:07 AM 
when should we evaluate left limit and right limit?  conjecture  Calculus  1  July 24th, 2008 01:14 PM 