Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :)

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

    Gib Z

    User Avatar
    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

    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

    User Avatar
    Homework Helper

    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:

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

    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....

    \lim_{n\to\infty} \frac{ e^{2n} + 1 }{ e^{n^2} + n }

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

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

    \lim_{n\to\infty} e^{2n-n^2} * 1

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook