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Operations on limit

by quasar987
Tags: limit, operations
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quasar987
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Nov16-04, 06:07 PM
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My Real Analysis textbook says: Let f,g: D --> R be two functions of common domain D that posses a limit at x_0 an accumulation point of D. Then, f/g as a limit at x_0 and this limit is the quotient of the limit of f to the limit of g, as long as [itex]g \neq 0 \ \forall x \ \epsilon \ D[/itex] and that the limit of g is not 0.

Does this mean that if the limit of g is zero we cannot conclude or could we extend the theorem to: if the limit of g is 0, then the limit of f/g does not exist?
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kreil
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Nov16-04, 08:31 PM
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(lim g) can't be 0 because otherwise (lim f/g) would be undefined
quasar987
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Nov16-04, 08:33 PM
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Is that a synonim of "the limit does not exist" ?

StatusX
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Nov16-04, 08:50 PM
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Operations on limit

I would say no, that the limit would be defined as infinity. for example, the limit as 1/x goes to 0 from the right is positive infinity, because 1 goes to 1 and x goes to 0. An example of an undefined limit would be the limit as x goes to 0 of sin(1/x).
HallsofIvy
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Nov17-04, 07:57 AM
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No. If g-> 0, then we the limit may or may not exist.

Obvious examples: Take f(x)= x, g(x)= x on the domain (0, 1). Then x= 0 is a an accumulation point. lim(x->0) f(x)= 0 and limit(x->0) g(x)= 0 so we can't use
(lim f(x))/(lim g(x)). But obviously f(x)/g(x)= 1 for all x in (0,1) so the limit as x-> 0 is just 1.

Take f(x)= x(x+a), g(x)= x on the domain (0,1). Again, g(x)-> 0 so we can't use
(lim f(x)/lim g(x)). But obviously f(x)/g(x)= x+a for all x in (0,1) so the limit as x-> 0 is just a. (The point of this example is that the "indeterminate" form 0/0 can give ANY number as limit.)

Take f(x)= x+ 1, g(x)= x on the domain (0,1). Again g(x)-> 0 so we can't use
(lim f(x)/lim g(x)). Here, for x close to 0, f(x) is close to 1 so we have f(x)/g(x)= 1/ very small number which give a very large number. There is no limit in this case.

(In general if g(x)->0 and f(x)-> non-zero number, there is no limit for f(x)/g(x). If BOTH f(x) and g(x) go to 0, then there may not be a limit or the limit may be any number.
StatusX
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Nov17-04, 01:41 PM
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maybe that's true, that limits with no bounds are said not to exist, but I think there is a big difference between a function that gets bigger and bigger in one direction as you move towards a point and one which stays finite, but does not settle on any value, such as sin(1/x) as x goes to 0.
HallsofIvy
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Nov19-04, 07:43 AM
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Yes, that's true. That's why some times we will say the limit "is infinity" or "is negative infinity" rather than just saying the limit does not exist. The limit STILL doesn't exist but you have a little more information about why it doesn't exist.
quasar987
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Nov22-04, 03:05 PM
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Thanks everyone for those insightful posts!


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