why doesn't lim x infinity (1+2/x)^x = infinity?

Discussion in 'Math' started by magnet18, Nov 3, 2012.

  1. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    Why?
    I found that the actual limit is e^2 using LHospitals rule, but why isn't it just infinity???
    From what I see, lim x → infinity (1+2/x)^x = (1+(2/∞))^∞ = (1+0)^∞ = ∞
     
  2. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    I'm assuming it's some rule about x being an exponent in a limit, or something like that
    possibly because you can't distribute the ^x to the 0, because 0^infinity is indeterminate?

    I would just like to know at what point in solving this i need to decide that i need to use e and Ln rather than simplification and plugging in

    exam tuesday :p
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    Why would (1+0)^∞ = ∞?

    What's 1^10? 1^100? 1^100000000?

    You don't know the 2/x is approaching zero "faster" than x is approaching infinity. If it is, then the answer will approach 1, if it isn't, then it will approach infinity, and if they two are "balanced" just right, it could converge on anything inbetween. In other words, it's indeterminate.

    y = (1+2/x)^x

    ln(y) = x * ln(1+2/x)

    As x=>∞, this goes to ln(y) => ∞*0 which is indeterminate, so we can use L'Hospital's rule.
     
  4. magnet18

    Thread Starter Senior Member

    Dec 22, 2010
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    OOOOOoooooooo...
    missed that bit
    *facepalm*
    thanks
     
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