Limit of quotient of two functions

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The discussion centers on the limit of the quotient of two functions, f(x) and g(x), with the claim that if the limit of f(x)/g(x) equals 1, it implies that the limits of f(x) and g(x) are equal. However, it is clarified that this implication does not necessarily hold true because the individual limits of f and g may not exist at the point where the limit of the quotient is evaluated. If the limits of f and g do exist, then they will indeed be equal. An example involving the limit of sin(x)/x illustrates this point, emphasizing that while the limit of the quotient can exist, it does not guarantee the existence of the individual limits. The conversation highlights the importance of understanding the conditions under which limits can be evaluated.
ModernLove
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Let f(x) and g(x) be functions.

Then if limit of f(x)/g(x) = 1. That implies lim f(x) = lim g(x) right?

Consider this proof.

lim f(x)/g(x) = 1
lim f(x) x lim 1/g(x) = 1
lim f(x) = 1 / (lim 1/g(x))
lim f(x) = lim g(x).
 
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No, the implication doesn't follow since the limits of f and g might not exist in the first place at the point where the limit of the quotient is 1.
 
Provided that the indivual limits actually exist then yes they will be equal. But just because the limit of f/g exists it doesn't mean that the limits of f and g neccessarily exist.


EDIT : No I'm not turning into a parrot, I must have posted the same time as arildno. :)
 
Last edited:
That's mind-reading, not parroting, uart.
 
arildno said:
That's mind-reading, not parroting, uart.

Or in this case just stating the obvious I think. :)
 
\lim_{x\rightarrow 1} \frac{\sin x}{x} = 1

\lim_{x\rightarrow 1} \sin x \neq 1

Can a mathematician clarify?
 
Eeh, you've got:
\lim_{x\to0}\frac{\sin(x)}{x}=1
 
Doh! I retract my previous statement.
 

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