My Math Forum A Proof concerning a negative f(x)

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 January 15th, 2018, 09:11 AM #1 Senior Member   Joined: May 2014 From: Allentown PA USA Posts: 107 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, 09:38 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,037 Thanks: 1063 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, 10:02 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,341 Thanks: 2463 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, 09:17 AM   #4
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Quote:
 Originally Posted by Carl James Mesaros 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
This makes no sense because you haven't said what L1 and L2 are! I suspect you mean that L1 is the limit of f(x) and L2 is the limit of g(x). But given that it should be lim f(x)g(x)= L1 times L2, not L1 minus L2.

Quote:
 (2) the limit as x --> infinity, f(x) = L (3) L is finite Prove: the limit as x --> infinity, [ -f(x) ] = -L Thank you.

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