Exploring Limits of arctan(ne)

[pivoxa15]

Homework Statement

Does lim(e->0)lim(n->infinity)arctan(ne)=pi/2?

and lim(n->infinity)lim(e->0)arctan(ne)=0?

If so why?

 

[Gib Z]

Are e and n independent variables? If so, then the limits don't actually exist.

 

[pivoxa15]

Yes they are independent. Why don't they exist?

 

[CompuChip]

I would say they do exist, basically
[tex]\lim_{\epsilon \to 0} \lim_{n \to \infty} \arctan n\epsilon = \lim_{\epsilon \to 0} \frac{\pi}{2} = \frac{\pi}{2}[/tex]
and similarly for the other one.
However it does show that
[tex]\lim_{(\epsilon, n) \to (0, \infty)} \arctan n\epsilon[/tex]
does not exist.

 

[HallsofIvy]

Gib z's point is this
[tex]\lim_{e\rightarrow 0}\left[\lim_{n\rightarrow \infty} arctan(ne)\right][/tex]
means "first take the limit as n goes to [/itex] infinity- then take the limit of that result as e goes to 0". The two limits are NOT independent. As n goes to infinity, the "inner" limit is [/itex]\pi/2[/itex] That is a constant so the "outer" limit, as e goes to 0, is also [itex]\pi/2[/itex].

On the other hand, if we take
[tex]\lim_{n\rightarrow \infty}\left[\lim_{e\rightarrow 0} arctan(ne)\right][/itex]
The inner limit is 0. That is a constant and so the outer limit is 0.

If the two limits are bo be taken "independently" (which most problems like this require) the limits as we approach in any way must exist and be the same. Since we have seen that two different ways of doing the two limits give different answers, the limit itself does not exist.

 

[pivoxa15]

That's an interesting point HallsofIvy. So the limit is undefined in both cases.

 

[Gib Z]

Well if you had put brackets in the right places, then your values in the first post would be correct. Without brackets, it implies that taking the limit in either order, or really at the same time, should get a single result, but the limit is multivalued.

Both cases really represent the same limit, e is an assigned pronumeral that could have been easily replaced with n, and vice versa for n. That is why I asked if the variables were independent.

 

[pivoxa15]

HallsofIvy had brackets and still claimed the limit doesn't exist.

 

[Avodyne]

HallsofIvy had brackets and still claimed the limit doesn't exist.

Actually, he said (correctly) that the limit does not exist if the two limits are to be taken "independently". If, on the other hand, you take the limits in a specified order, or in some other specified way (such as [itex]n\to\infty[/itex] and [itex]\epsilon\to0[/itex] with [itex]n\epsilon[/itex] fixed), then the limit does exist. In physics, when this situation arises, people often say that there is "an order-of-limits problem", or that "the limits don't commute".

 

[pivoxa15]

So if I say, one must take [tex]\lim_{e\rightarrow 0}\left[\lim_{n\rightarrow \infty} arctan(ne)\right][/tex] and only that one, in other words take only n as it goes to infinity first then the other then it makes sense?

So with this condition, the limits are now dependent?

 

[Gib Z]

Well for the example you gave, you do them just like other things, work from the inside out.

So the thing inside the bracket, Where the limit if where n --> infinity, the "e" part is treated as a constant, then we evaluate the part inside to brackets to be pi/2, and then we take the limit as e-->0, but there is no e left in pi/2. The confusion with the order of limits arises when one considers the limit operator as a separate entity. Remember that we must always consider the expression it is connected to. eg I am sure you wouldn't confuse [tex](\sqrt{x})^2[/tex] and [tex]\sqrt{x^2}[/tex], which are 2 different functions. Changing the order of operators around is generally not allowed.

 

[HallsofIvy]

The point was deciding exactly what the problem asked. Since the function here, arctan(ne) is exactly the same, taken as a "two dimensional limit", both would be the same: "limit does not exist". However, I suspect that here the problem was to take the two limits in the order given-i.e. to show that the two limits are different.

 

[pivoxa15]

The problem was actually an integration with n->infinity presupposed. I later introduced e->0 into the integration to make it doable so the limit e->0 is meant to be taken after n so in this case everything works and the limit exists and is pi/2. In other words in this problem the limits are not independent after all.

 

[HallsofIvy]

It that case I think it would be a good idea to show us the original problem so we can try to understand what you are saying!

 

[pivoxa15]

The problem reduces to show the limit of lim(e->0)lim(n->infinity)arctan(ne) and only that. So order of taking the limit is important. That limit exists and equals pi/2, correct?

 

[Gib Z]

yes its correct.