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Homework Help: Limit n approaches infinity of (1 + n)^(1/n)

  1. Apr 9, 2013 #1
    1. lim (1 + n)^(1/n)

    2. I was able to figure out that the limit goes to 1 only after I substituted larger and larger values in place of n in my calculator.

    Since I cannot use a calculator on my test, is there another way to know what the limit goes to?

  2. jcsd
  3. Apr 9, 2013 #2

    Ray Vickson

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    Take logarithms.
  4. Apr 9, 2013 #3
    AGH! Taking logarithms seems harder than it should be...

    Just setting up a log would look like this, right? log1+nx = 1/n

    ...Don't even know how to solve that.

    But isn't there an intuitive way to figure out the right answer?
  5. Apr 9, 2013 #4


    Staff: Mentor

    Not a very convenient log base. Try loge (AKA ln).
  6. Apr 9, 2013 #5
    I would start approaching this by considering that: lim_(x→∞) for (1/x) = 0
  7. Apr 9, 2013 #6


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    Are you sure you wrote the limit correctly? There's a much more interesting limit that goes ##\displaystyle \lim_{n \to \infty} (1 + \frac{1}{n})^n## or, equivalently, ##\displaystyle \lim_{n \to 0} (1 + n)^{\frac{1}{n}}##.

    As written, the limit is easily seen to be equivalent to ##\displaystyle \lim_{n \to \infty} n^{\frac{1}{n}}## since the "1" can be ignored at the limit. This limit can be resolved by taking the natural logarithm, and then using L' Hopital's Rule.
  8. Apr 9, 2013 #7
    Okay, thanks, you guys! :)

    So I take the ln to bring down the fraction.


    = 1/2ln(n+1) →L'Hospital's→ (1/2)(1/(n+1))(1) = 1/(2n + 2)

    When I substitute infinity for n here, it seems that it goes to 0, not 1...

    Am I doing something wrong here...?
  9. Apr 9, 2013 #8


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    Where did the (1/2) power come from? And if the log goes to 0 what does the original sequence go to?
  10. Apr 10, 2013 #9
    oh, man. I was looking at another problem and wrote down 1/2 instead of 1/n!

    ln(1+n)1/n = ln(1+n)/n →L'Hospital's→ 1/(1+n)

    1/(1+infinity) = 0

    Oh, so you mean that since we used the ln, and it went to zero, then the original expression must have gone to 1.

    Because ln(1) = 0

    Is this right?

    Thanks! :)
  11. Apr 10, 2013 #10


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    Pretty much right, except that I think you should use limit notation instead of writing things like 1/(1+infinity) = 0.
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