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Lim as n->infinity help

  1. Jun 17, 2005 #1
    It seems to an area I'm really fuzzy on. Anyway, the question I've having problems with is

    lim as n->infinity[tex]
    (1 + \frac{1}{n+1}) ^ {3-n}

    At first, I thought I could evaluate the stuff inside the bracket first and then take care of the power ie

    lim as n->infinity[tex]
    (1 + \frac{1}{n+1})
    [/tex] = 1

    so lim as n->infinity[tex]
    (1 + \frac{1}{n+1}) ^ {3-n} = 1 ^ {3-n} = 1

    But that's not right... (I drew up a graph and did some calculations, it's supposed to be 0.36-something)

    Then I thought I could make it equal to
    [tex]e^{lim (3-n)ln\frac{n+2}{n+1}}[/tex]
    so I could get rid of the (3-n) power, but then I would end up with infinity * 0 on top of the e.

    A clue as to what I should be doing please? ^^ Thanks.
  2. jcsd
  3. Jun 18, 2005 #2


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    Homework Helper

    First I would change the variable from n to m-1. If m goes to infinity, n does as well, so you'll get the same limit. Then, do you know the formula for e? See if you can put this in a similar form.
  4. Jun 18, 2005 #3
    use exponent rules to reduce the original expression. [tex] X^{a+b}= X^a X^b [/tex] and then try it.
  5. Jun 18, 2005 #4
    Ok using what both of you said...

    if n = m-1

    [tex](1 + \frac{1}{m}) ^ {2-m}[/tex]

    and using an exponent rule to get

    [tex]\frac {(1 + \frac{1}{m}) ^ 2}{(1 + \frac{1}{m}) ^ m}[/tex]
    [tex]= \frac {1}{e}[/tex]

    ... which seems to be the answer... thanks very much StatusX and Gale17 :)
  6. Jun 18, 2005 #5
    oops sorry, i'm missing the lim as m->infinity signs...

    well if you put those in, it makes sense :)
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