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# Analysis Limits Homework Problem watch

1. Hey guys, so I'm sort of iffy on whether or not my answer to this question is correct so I was just wondering if someone could quickly glance over it. In particular im not so sure on the part about letting but I couldnt really think of any other way to show it.

Question and attempt are (hopefully) attached.

Thanks
Attached Images

2. (Original post by DylanJ42)
..
This proof isn't valid. Your new choice for epsilon (i.e. ) must be a constant - it cannot depend on n.

Hint: The result you show in brackets is going to be more helpful to you than "we have x_n having the same sign as x*".
3. (Original post by DFranklin)
This proof isn't valid. Your new choice for epsilon (i.e. ) must be a constant - it cannot depend on n.

Hint: The result you show in brackets is going to be more helpful to you than "we have x_n having the same sign as x*".
ah okay thank you, Ill give it another go now
4. (Original post by DFranklin)
This proof isn't valid. Your new choice for epsilon (i.e. ) must be a constant - it cannot depend on n.

Hint: The result you show in brackets is going to be more helpful to you than "we have x_n having the same sign as x*".
so I think I have something but it relies on epsilon being less than x*, also the negative and positive cases for x* had to be considered separately. Does this sound along the right lines or am I miles off?
5. (Original post by DylanJ42)
so I think I have something but it relies on epsilon being less than x*, also the negative and positive cases for x* had to be considered separately. Does this sound along the right lines or am I miles off?
This sounds too complicated to be correct, although it may well be.

You have ; you can control the size of the numerator as much as you want since you know that , so it can be made arbitrarily small. All you need to do is show that the denominator is bounded. You can now either (as DFrank) mentions use something akin to what you have in your brackets or use a fairly elementary theorem about convergent sequences and boundedness. I'll leave you to it.
6. (Original post by Zacken)
x
ok so I understand your second suggestion, although im still sort of confused on DFranklins hint,

is the idea to show that for all n>= 1 there's an epsilon which satisfies |xn - x*|< E therefore the difference in value of xn and x* is finite therefore xn is bounded?
7. (Original post by DylanJ42)
ok so I understand your second suggestion, although im still sort of confused on DFranklins hint,

is the idea to show that for all n>= 1 there's an epsilon which satisfies |xn - x*|< E therefore the difference in value of xn and x* is finite therefore xn is bounded?
No. The big issue you have in trying to show 1/x_n is bounded is that everything blows up if x_n can get arbitrarily close to 0. So part of what you want to show that there's a bound m such that for all sufficiently large n.
8. (Original post by DFranklin)
No. The big issue you have in trying to show 1/x_n is bounded is that everything blows up if x_n can get arbitrarily close to 0. So part of what you want to show that there's a bound m such that for all sufficiently large n.
something really isnt clicking here sorry

i get that showing x_n > m for large enough n tells us that 1/x_n isnt going to go off to infinity, but yea im not really getting where you're going with that.

Doesn't the question saying that x_n -> x* and x* =/= 0 just imply that x_n doesn't get arbitrarily close to 0 since it must be getting very very close to some non zero x*?
9. (Original post by DylanJ42)
something really isnt clicking here sorry

i get that showing x_n > m for large enough n tells us that 1/x_n isnt going to go off to infinity, but yea im not really getting where you're going with that.

Doesn't the question saying that x_n -> x* and x* =/= 0 just imply that x_n doesn't get arbitrarily close to 0 since it must be getting very very close to some non zero x*?
Yes, but you do need to actually put that into proper mathematical language.
10. (Original post by DFranklin)
Yes, but you do need to actually put that into proper mathematical language.
oh okay i see I didnt realise it had to be explicitly shown, ill try my best from there and hopefully get it.

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