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Challenged in evaluating this limit.

  1. Apr 16, 2007 #1
    I really haven't got a clue on how to evaluate this limit. I've tried doing algebraic manipulation, but to no avail. (L'Hopital's rule are not allowed to be used). If someone can give me a hint, that would be great :)

    [tex]
    \lim_{n\to\infty} \frac{(e^n)^2 + 1}{e^{n^2}+n}
    [/tex]
     
  2. jcsd
  3. Apr 16, 2007 #2

    Gib Z

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    Homework Helper

    I don't understand your notation on the denominator, is that e^(n^2) or what?
     
  4. Apr 16, 2007 #3
    i am not quite sure on this, however i think it should go like this:

    substitute e^n=t, than you get, ln e^n=lnt=>n=ln t, when n->infinity, t->infinity

    lim(t^2+1)/(t^2+ln t),t->infinity

    lim_t->infinity(t^2)/(t^2+lnt)+lim_t->infinity(1)/(t^2+lnt)
    the second part is obviously 0 when t-> infinity, so we are left with

    lim_t->infinity(t^2)/(t^2+lnt), we know that ln t=ln(1+(t-1)), which is equivalent with t-1, because the limit of their ratio is 1, we substitute it and we get

    lim_t->infinity(t^2)/(t^2+t-1), so the limit of this is obviously 1.
     
  5. Apr 16, 2007 #4

    Gib Z

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    Oh ok I see your latex so I know what you mean. Give me a second to see if I can work it.
     
  6. Apr 16, 2007 #5
    the denominator i enterpreted as (e^n)^2,
    I do not know if this is what he meant????

    Is this what you meant???
     
    Last edited: Apr 16, 2007
  7. Apr 16, 2007 #6
    I meant e^(n^2). So yeah, obviously the one you did with the 't' substitution is not what I'm looking for.
     
    Last edited: Apr 16, 2007
  8. Apr 16, 2007 #7
    than i guess the limit of that is going to be 0, since the denominator growes much faster than the numerator.
    I will see if i can work it out now
     
  9. Apr 16, 2007 #8
    try it with the same substitution e^n=t, but just work it out in the form u meant. e^(ln^2 t), take this substitution then for e^(n^2), and i think you will get what u are looking for!
     
    Last edited: Apr 16, 2007
  10. Apr 16, 2007 #9
    Ok, so then you get:

    [tex]
    \lim_{t\to\infty} \frac{t^2 + 1}{e^{(\ln{t})^2} + \ln{t}}
    [/tex]

    So basically you will get infinity at the numerator and denominator, which doesn't lead to anything as far as I can see... *sigh*
     
    Last edited: Apr 16, 2007
  11. Apr 16, 2007 #10
    Denominator grows WAY faster than the numerator, so in this case it goes to 0
     
  12. Apr 16, 2007 #11
    Can you just say that without doing any calculations? o_O
     
  13. Apr 16, 2007 #12

    i guess we cannot just say this, if this would be in an exam. however it is a clue to lead us on the right direction. i think we should do some calculations.

    Although my approach i guess is obviously not appropriate(to much calculations).
     
  14. Apr 16, 2007 #13
    I think I just worked it out....

    [tex]
    \lim_{n\to\infty} \frac{ e^{2n} + 1 }{ e^{n^2} + n }
    [/tex]

    [tex]
    \lim_{n\to\infty}\frac{ e^{2n}(1 + e^{-2n}) }{ e^{n^2}(1 + ne^{-n^2}) }
    [/tex]

    [tex]
    \lim_{n\to\infty} e^{2n-n^2} \frac {(1 + e^{-2n}) }{(1 + ne^{-n^2}) }
    [/tex]

    [tex]
    \lim_{n\to\infty} e^{2n-n^2} * 1
    [/tex]

    So that just equals to 0.
     
    Last edited: Apr 16, 2007
  15. Apr 16, 2007 #14
    how do you know that the limit of the second part is 1
     
  16. Apr 16, 2007 #15
    Yeah, the ne^(-n^2) part is actually kind of part of the question. In the question it gives you limits of various functions - I didn't list it here because there is too many. So yeah, basically that term goes to 0 as n approaches infinite.
     
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