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    Prove that, as  n \rightarrow \infty , the limit of the sequence  u_{n} = \frac{1}{n^2+1} is 0.

    So,

    Let  \varepsilon > 0 be given.

     | u_{n} - 0 | < \varepsilon
    iff  \frac{1}{n^2+1} < \varepsilon
    iff  m^2 + 1 > \frac{1}{\varepsilon}
    iff  m^2 > \frac{1}{\varepsilon} - 1 *
    iff  m > \sqrt{max(\frac{1}{\varepsilon}-1, 0)} **

    I don't understand the last two step, so
    i) Is it that we are really only interested in small positive  \varepsilon , rather than a big value or 10 or something?
    ii) Why the last step? I mean why must m bigger than the square root of the larger of the two values there? I'm confused.

    Thanks.
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    (Original post by melbss01)
    Prove that, as  n \rightarrow \infty , the limit of the sequence  u_{n} = \frac{1}{n^2+1} is 0.

    So,

    Let  \varepsilon > 0 be given.

     | u_{n} - 0 | < \varepsilon
    iff  \frac{1}{n^2+1} < \varepsilon
    iff  m^2 + 1 > \frac{1}{\varepsilon}
    iff  m^2 > \frac{1}{\varepsilon} - 1 *
    iff  m > \sqrt{max(\frac{1}{\varepsilon}-1, 0)} **

    I don't understand the last two step, so
    i) Is it that we are really only interested in small positive  \varepsilon , rather than a big value or 10 or something?
    ii) Why the last step? I mean why must m bigger than the square root of the larger of the two values there? I'm confused.

    Thanks.
    the last step is because if you chose epsilon like 10 then you have the root of a negative which isn't going to be very useful here.
    for the other step, we're interested in all values of epsilon, the idea behind these proofs is to find an m in terms of epsilon that will work for any epsilon.
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    (Original post by Totally Tom)
    the last step is because if you chose epsilon like 10 then you have the root of a negative which isn't going to be very useful here.
    for the other step, we're interested in all values of epsilon, the idea behind these proofs is to find an m in terms of epsilon that will work for any epsilon.
    Thanks. I have a few more questions.

    1) Why is it not going to be very useful that  m^2 is greater than  a , where  a is negative?

    2) For the last step, why do we want the larger value of  \frac{1}{\varepsilon} and 0?

    Thanks.
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    (Original post by melbss01)
    Thanks. I have a few more questions.

    1) Why is it not going to be very useful that  m^2 is greater than  a , where  a is negative?

    2) For the last step, why do we want the larger value of  \frac{1}{\varepsilon} and 0?

    Thanks.
    1.) what is a. if you're referring to my point about what is epsilon was 10 or w/e just think about it. you'd have sqrt(-0.9) now, how are you going to relate that in any way to m^2.
    2.) i explained this, it's not the larger value of 1/epislon and 0 it's the larger value of 1/epsilon MINUS 1 and 0 that's why you want the max value of the
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    Same reason for both: you're getting a formula for (a lower bound for) m in terms of epsilon. If your formula involves the square root of a negative number for some values of epsilon, then it's not a very good formula, is it?
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    (Original post by Totally Tom)
    1.) what is a. if you're referring to my point about what is epsilon was 10 or w/e just think about it. you'd have sqrt(-0.9) now, how are you going to relate that in any way to m^2.
    2.) i explained this, it's not the larger value of 1/epislon and 0 it's the larger value of 1/epsilon MINUS 1 and 0 that's why you want the max value of the
    1) I'm refering "a" as a negative number, happens when some values of epsilon such that  \frac{1}{\varepsilon} - 1 is negative.

    2) Sorry but I really forgot to type a '1' there. Anyway, so you want RHS to be positive, that's why you choose the larger value of  \frac{1}{\varepsilon} -1 (Let this be y) and 0? I understand that that you would choose 0 when y is negative. My question is, when y is more than 0, why do you choose y instead of 0? Sorry for such question but I really need to clarify this point.

    (Original post by DFranklin)
    Same reason for both: you're getting a formula for (a lower bound for) m in terms of epsilon. If your formula involves the square root of a negative number for some values of epsilon, then it's not a very good formula, is it?
    Yes, thanks. Another question though,

    when we solve
     m^2 > \frac{1}{\varepsilon} - 1 , shouldn't it be

     |m| > \sqrt{max(\frac{1}{\varepsilon}-1,0)} ?

    I'm referring to the usual inequality we did.

     x^2 > 3
    iff  |x| > \sqrt{3} .

    Thanks.
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    No, because by definition you have m >= 0.
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    (Original post by DFranklin)
    No, because by definition you have m >= 0.
    Sorry I typed "m" instead of "n".

    The definition you mean is the precise definition of the limit of sequence?

    Or

    Is it because of the term "as n goes to infinity", so n is greater than 0?
    How about when n goes to negative infinity, does it make any difference? (I know the limit is still the same)

    Thanks.
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    In normal terminology, you have a sequence (a_n) = a_1, a_2, ... so it makes no sense to talk about n < 0.

    If you want to consider what happens as n \to -\infty, you basically have to rework your definition of limit to deal with n < 0. It's obvious(*) what the definition ought to be, but (I think) at the end of the day it's still a definition. The most common thing to do would be to define a sequence (b_n) = (a_{-n}) (for n = 1, 2, 3, ...) and then consider
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \lim_{n \to \infty) b_n
    .

    (*) Despite the impression you might get from lecturers, you do actually need to use common sense for analysis.
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    (Original post by DFranklin)
    In normal terminology, you have a sequence (a_n) = a_1, a_2, ... so it makes no sense to talk about n < 0.

    If you want to consider what happens as n \to -\infty, you basically have to rework your definition of limit to deal with n < 0. It's obvious(*) what the definition ought to be, but (I think) at the end of the day it's still a definition. The most common thing to do would be to define a sequence (b_n) = (a_{-n}) (for n = 1, 2, 3, ...) and then consider
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \lim_{n \to \infty) b_n
    .

    (*) Despite the impression you might get from lecturers, you do actually need to use common sense for analysis.
    Thanks.
 
 
 
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