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Finding a limit

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm having trouble with these here.. it's been a while since I've done sequences and I can't seem to make this work with Standard Limits equations.

    Clearly the answer given by Wolfram solver is there after the = but i'd like to know the reasoning behind it.

    Anyone that could point me in the right direction would be most helpful!

    1. zuf9eg.png

    2. 5plz8.png

    3. anbz1x.png


    thanks a lot

    -chief10
     
    Last edited: Dec 4, 2012
  2. jcsd
  3. Dec 4, 2012 #2

    haruspex

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    Do you know the standard limit result for (1+1/n)n? Yo can use that to solve 1 and 3.
     
  4. Dec 4, 2012 #3
    All you need to know is that [itex] e^x = \lim_{n\rightarrow \infty} (1 + \frac{x}{n})^{n} [/itex] and that if [itex] \lim f(x) = a [/itex] and [itex] \lim g(x) = b [/itex] with a,b≠∞, then [itex] \lim f(x) g(x) = a b [/itex]
     
  5. Dec 4, 2012 #4
    so you just divide inside the brackets by 'n' right?

    i don't get how the (n+1) part works with the particular standard limit though.. it's throwing me off a bit..

    also what do you do with n^n..
     
  6. Dec 4, 2012 #5
    ah yeah just your standard limit laws

    how do you apply it to this one though..
     
    Last edited: Dec 4, 2012
  7. Dec 4, 2012 #6
    I couldn't work it out

    SL 7 didn't work for me.. any more tips?
     
  8. Dec 4, 2012 #7

    haruspex

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    Yes. Please post your working as far as you get.
     
  9. Dec 4, 2012 #8

    dextercioby

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    You can also use the definition of e in terms of a limit for point 2, so you can have a common starting point for all 3. But using e for point 2 might be an overkill, though.
     
  10. Dec 4, 2012 #9
    Actually for the third one you don't need the definition of e.

    0≤n[itex]^{n}[/itex]/(3+n)[itex]^{n+1}[/itex]≤ ?? which approaches 0.
     
  11. Dec 5, 2012 #10
    This is what I've done for (1-(1/(n^2)))^n ===> I think it works

    exp[lim(n→∞)n*log(1-(1/(n^2)))

    let t = 1/n and use L'Hopitals

    exp[2*lim(t→0)(t/((t^2)-1))

    then take the limit of each individual component inside the exponent and you should be left with exp(2*0/0-1) = exp(0) = 1

    sound about right?

    reckon i should do the same thing [exp(log)] type approach for the others?
     
  12. Dec 5, 2012 #11
    could you elaborate further? i've never seen that concept before
     
  13. Dec 5, 2012 #12
    for number 1 i can't seem to solve it using exp[log]...

    L'Hopitals doesn't work since the denominator is 1..
     
  14. Dec 5, 2012 #13
    Number 1 might be conceptually easier if you first write m = n+2 and consider m→∞. You can split it into two fractions, after which the rule I mentioned earlier becomes useful.
     
  15. Dec 5, 2012 #14
    You know that the sequence is positive, if you show that the sequence is bounded by another sequence and that bounded sequence approaches 0, then you can deduce that your original sequence is 0. Similar concept to squeeze theorem.
     
  16. Dec 5, 2012 #15
    For the second one you can do in 1 step. Factor it then apply the definition of e^x and you get 1/e * e = 1.
     
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