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Limit q help

I need help on 2iv and 2v. I had a go at 2iv but im not sure if it is correct, seemed too easy. And for 2v, I have no idea but it does seem like the limit doesn't exist although I really don't know how to prove it.

Can someone help please.

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Reply 1
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
I need help on 2iv and 2v. I had a go at 2iv but im not sure if it is correct, seemed too easy. And for 2v, I have no idea but it does seem like the limit doesn't exist although I really don't know how to prove it.

Can someone help please.

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Sorry cooldude.

I wish I could help but I can only do "usual" limits ...

it looks that all the purists have gone to bed early or all out partying ...
Reply 2
Avatar for james22
james22
16
iv seems fine to me. It's a question meant to test whether you can notice that the trig terms don't matter at all, since they are bounded. x*(any complicated but bounded function) tends to 0 as x tends to 0.

For v, you are right that the limit doesn't exist. Proving it relies on the fact that every intervals of the real numbers contains both rational and irrational numbers.
Original post by TeeEm
Sorry cooldude.

I wish I could help but I can only do "usual" limits ...

it looks that all the purists have gone to bed early or all out partying ...


Cant really blame them since its friday. And dw im sure I'll have a lot more multivariable questions to ask lol.

Original post by james22
iv seems fine to me. It's a question meant to test whether you can notice that the trig terms don't matter at all, since they are bounded. x*(any complicated but bounded function) tends to 0 as x tends to 0.

For v, you are right that the limit doesn't exist. Proving it relies on the fact that every intervals of the real numbers contains both rational and irrational numbers.


Im not really understanding... about v. Can you elaborate a bit more


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Reply 4
Avatar for james22
james22
16
Original post by cooldudeman
Cant really blame them since its friday. And dw im sure I'll have a lot more multivariable questions to ask lol.



Im not really understanding... about v. Can you elaborate a bit more


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To prove that a limit does not exist, a good way to start is to assume that it does and use the uniqueness of limits to derive a contradiction. Let the limit be x_0 and see what you can deduce from that.
Original post by james22
To prove that a limit does not exist, a good way to start is to assume that it does and use the uniqueness of limits to derive a contradiction. Let the limit be x_0 and see what you can deduce from that.


I dont get it because
Lets assume that the limit is x_0 when x tends to zero. Zero is a rational number so fx is going to be 1-x=x_0=1-0=1.

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Reply 6
Avatar for james22
james22
16
Original post by cooldudeman
I dont get it because
Lets assume that the limit is x_0 when x tends to zero. Zero is a rational number so fx is going to be 1-x=x_0=1-0=1.

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When finding the limit you are not at all concerned about what value the function takes at 0, but what values it takes around 0.
Original post by james22
When finding the limit you are not at all concerned about what value the function takes at 0, but what values it takes around 0.

Is this fully correct?

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Reply 8
Avatar for james22
james22
16
Original post by cooldudeman
Is this fully correct?

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Mostly fine. When you say any sequence {x_n}, this needs to be any sequence which only takes the value 0 finitely often, and converges to 0. Also on the last line you mention continuty, but you are trying to prove that the limit doesn't exist. A function can be discontinuous at a point but still have a limit there.
Original post by james22
Mostly fine. When you say any sequence {x_n}, this needs to be any sequence which only takes the value 0 finitely often, and converges to 0. Also on the last line you mention continuty, but you are trying to prove that the limit doesn't exist. A function can be discontinuous at a point but still have a limit there.


Can you explain more with the part when you said:
When you say any sequence {x_n}, this needs to be any sequence which only takes the value 0 finitely often, and converges to 0.

Finitely often?
Original post by cooldudeman
Can you explain more with the part when you said:
When you say any sequence {x_n}, this needs to be any sequence which only takes the value 0 finitely often, and converges to 0.

Finitely often?


Consider the condition which is 0 everywhere apart from at 0, where it is 1. The limit of this function as x->0 is 0, yet I could choose the sequence x_n=0 which would suggest that the limit was 1 if I did not have the finiteness condition. Have you covered the links between limits of functions and sequences yet? The finiteness condition should be part of the theorem saying that they are equivalent.
Original post by james22
Consider the condition which is 0 everywhere apart from at 0, where it is 1. The limit of this function as x->0 is 0, yet I could choose the sequence x_n=0 which would suggest that the limit was 1 if I did not have the finiteness condition. Have you covered the links between limits of functions and sequences yet? The finiteness condition should be part of the theorem saying that they are equivalent.


We have covered functions linking with sequences and how the limit if all the sequences have to tend to the same limit but the word finiteness never came in.

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Original post by cooldudeman
We have covered functions linking with sequences and how the limit if all the sequences have to tend to the same limit but the word finiteness never came in.

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Can you post a copy of the theorem? There are a couple of other ways of stating it which all work.
Original post by james22
Can you post a copy of the theorem? There are a couple of other ways of stating it which all work.


Tbh ours doesn't really have an introduction to the proof.

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Original post by cooldudeman
Tbh ours doesn't really have an introduction to the proof.

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OK, the key line in that then is the condition that a_n=/=a. It is actually OK to allow a_n=a so long as it only happens finitely often.
Original post by james22
OK, the key line in that then is the condition that a_n=/=a. It is actually OK to allow a_n=a so long as it only happens finitely often.


Ok judging from my definition, is this all correct? Have I missed out anything?

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Original post by cooldudeman
Ok judging from my definition, is this all correct? Have I missed out anything?

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Looks good.

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