January 15th, 2018, 10:11 AM  #1 
Senior Member Joined: May 2014 From: Allentown PA USA Posts: 104 Thanks: 6 Math Focus: dynamical systen theory  A Proof concerning a negative f(x)
Dear My Math Forum Community: What would be a suitable proof using the information below? Given: (1) the limit as x > infinity, f(x)g(x) = L1  L2 (2) the limit as x > infinity, f(x) = L (3) L is finite Prove: the limit as x > infinity, [ f(x) ] = L Thank you. 
January 15th, 2018, 10:38 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,755 Thanks: 900 
we are told that $\lim \limits_{x\to \infty} ~f(x) = L,~L \text{ is finite}$ $1 \text{ is a finite constant}$ therefore by the constant law of limits $\lim \limits_{x \to \infty}~{f(x)} = \lim \limits_{x \to \infty}~(1) f(x) = (1) \lim \limits_{x \to \infty}~ f(x) = (1) L = L$ 
January 15th, 2018, 11:02 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,138 Thanks: 2381 Math Focus: Mainly analysis and algebra 
I assume there's an error in point 1) and that using $g(x)=1$ gives the required result.

January 20th, 2018, 10:17 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,943 Thanks: 797  Quote:
Quote:
 

Tags 
negative, proof 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proof that i is poth positive and negative  Tau  Math  40  December 31st, 2016 07:23 PM 
power rule Proof for negative Exponents  maxgeo  Calculus  1  June 10th, 2014 02:48 PM 
Manipulation of negative square root of a negative term/#  daigo  Algebra  3  June 30th, 2012 08:06 AM 
Negative root Proof  jstarks4444  Number Theory  11  February 17th, 2011 04:48 PM 
just can't be negative, right?  empiricus  Algebra  6  May 4th, 2010 02:30 PM 