Limits of f(x) & g(x) Do Not Equal Lim[f(x)g(x)]

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SUMMARY

The discussion centers on the misconception that if the limits of two functions, f(x) and g(x), both exist and are equal as x approaches a, then the limit of their product, Lim[f(x)g(x)], equals 1. Participants clarify that this statement is false, particularly when considering cases where the limits approach ±∞. The correct interpretation involves recognizing that the notation Lim[x→a] indicates a two-sided limit, and that division rather than multiplication may be the relevant operation in certain contexts.

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Homework Statement


If Lim f(x) and Lim g(x) both exist and are equal
x→a x→a then Lim[f(x)g(x)]=1
x→a

Homework Equations


No relevant equations are required in this problem. To determine whether the statement is true or false [/B]

The Attempt at a Solution


The statement is false but the reason behind it is quite unclear to me. Is it because when the limit approaches a from left and right the limits are -∞ and +∞?? Would be very helpful if anyone of you could explain it.
 
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Hard to read your post. Something like

If ##\displaystyle \lim_{x\rightarrow a} f(x)## and ##\displaystyle \lim_{x\rightarrow a} g(x)## both exist and are equal ##\Rightarrow \displaystyle \lim_{x\rightarrow a} f(x)g(x) = 1 ## ?

Could that be ##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ## ?

  • To me ##\displaystyle \lim_{x\rightarrow a}## means "x approaching from the left to a" so I wouldn't worry about "coming from the right"
  • To me ##\displaystyle \lim_{x\rightarrow a} f(x) = \pm \infty## means the limit does not exist, so I wouldn't worry about those either
  • But the division cause a problem in a particular case, that can serve as a counter-example that makes the general statement not true,
 
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BvU said:
Hard to read your post. Something like

If ##\displaystyle \lim_{x\rightarrow a} f(x)## and ##\displaystyle \lim_{x\rightarrow a} g(x)## both exist and are equal ##\Rightarrow \displaystyle \lim_{x\rightarrow a} f(x)g(x) = 1 ## ?

Could that be ##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ## ?

  • To me ##\displaystyle \lim_{x\rightarrow a}## means "x approaching from the left to a" so I wouldn't worry about "coming from the right"
The notation ##\lim_{x \to a} f(x)## indicates a two-sided limit in which x can approach a from either side. If the two-sided limit exists, then both one-sided limits also exist. Maybe that's what you were trying to say, but what you actually said wasn't clear.
BvU said:
  • To me ##\displaystyle \lim_{x\rightarrow a} fx) = \pm \infty## means the limit does not exist, so I wouldn't worry about those either
  • But the division cause a problem in a particular case, that can serve as a counter-example that makes the general statement not true,
I agree that the first post was unclear, and that the OP meant division instead of multiplication.
 
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BvU said:
Hard to read your post. Something like

If ##\displaystyle \lim_{x\rightarrow a} f(x)## and ##\displaystyle \lim_{x\rightarrow a} g(x)## both exist and are equal ##\Rightarrow \displaystyle \lim_{x\rightarrow a} f(x)g(x) = 1 ## ?

Could that be ##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ## ?

  • To me ##\displaystyle \lim_{x\rightarrow a}## means "x approaching from the left to a" so I wouldn't worry about "coming from the right"
  • To me ##\displaystyle \lim_{x\rightarrow a} fx) = \pm \infty## means the limit does not exist, so I wouldn't worry about those either
  • But the division cause a problem in a particular case, that can serve as a counter-example that makes the general statement not true,

Standard usage is that ##x \to a## means a two-sided limit (##|x-a| \to 0##). Left-hand limits are typically denoted as ##x \to a\!-## or ##x \to a\!-\!0## or ##x \uparrow a##. Right-hand limits are typically denoted as ##x \to a\!+## or ##x \to a\!+\!0## or ##x \downarrow a##.
 
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Good thing you corrected me, gentlemen. Been too long ago, but I do recognize the ##x\downarrow a## and ##x\uparrow a\;.\ ## Never had much opportunity to make good use of the distinction. Apologies to professor D.
 

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