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January 15th, 2018, 09:11 AM   #1
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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.
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January 15th, 2018, 09:38 AM   #2
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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

$\lim \limits_{x \to \infty}~{-f(x)} = \lim \limits_{x \to \infty}~(-1) f(x) = (-1) \lim \limits_{x \to \infty}~ f(x) = (-1) L = -L$
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January 15th, 2018, 10:02 AM   #3
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I assume there's an error in point 1) and that using $g(x)=-1$ gives the required result.
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January 20th, 2018, 09:17 AM   #4
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Quote:
Originally Posted by Carl James Mesaros View Post
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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