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Find the limit

  1. Oct 16, 2004 #1

    quasar987

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    [tex]\lim_{n\rightarrow \infty} \frac{2^{3n}}{3^{2n}}[/tex]

    The answer is zero. All I can do is turn this infinity/infinity undeterminate form into a 0 times infinity indeterminate form. I also tried finding a creature strictly bigger than [itex]\frac{2^{3n}}{3^{2n}}[/tex] that has zero for a limit so that the answer would follow from the "sandwich theorem". But all my attempts let to infinity. For instance,

    [tex]0\leq \frac{2^{3n}}{3^{2n}}\leq \frac{3^{3n}}{3^{2n}}=\frac{3^{2n}3^n}{3^{2n}}=3^n[/tex]
     
    Last edited: Oct 16, 2004
  2. jcsd
  3. Oct 16, 2004 #2
    2^(3n) = (2^3)^n = 8^n and 3^(2n) = (3^2)^n = 9^n, so 2^(3n) / 3^(2n) = what?
     
  4. Oct 16, 2004 #3

    quasar987

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    [tex]\lim_{n\rightarrow \infty} \left( \frac{8}{9} \right)^n=0[/tex]

    Thanx a bunch!
     
    Last edited: Oct 16, 2004
  5. Oct 16, 2004 #4
    Btw, [tex]\infty[/tex] is written \infty in LaTeX.
     
  6. Oct 16, 2004 #5

    quasar987

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    I would have another question for you if you don't mind.

    The textbook says that

    [tex]\lim_{n\rightarrow \infty} \sqrt{2n+1} - \sqrt{2n}=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{2n+1}+\sqrt{2n}}[/tex]

    What are the steps to get from one to the other?
     
    Last edited: Oct 16, 2004
  7. Oct 16, 2004 #6
    Multiply [tex]\sqrt{2n + 1} - \sqrt{2n}[/tex] by its conjugate like so:

    [tex]\sqrt{2n + 1} - \sqrt{2n} = (\sqrt{2n + 1} - \sqrt{2n}) \cdot \frac{\sqrt{2n + 1} + \sqrt{2n}}{\sqrt{2n + 1} + \sqrt{2n}}[/tex].

    The top can then be simplified into 1...
     
  8. Oct 16, 2004 #7

    quasar987

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    Great!

    Those were #1 a) and b) out of n) by the way. Wish me luck!
     
  9. Oct 16, 2004 #8
    Is this from Courant's book?
     
  10. Oct 16, 2004 #9

    quasar987

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    No, it's from two Canadian people: Jaques Labelle and Armel Mercier.
     
    Last edited: Oct 16, 2004
  11. Oct 18, 2004 #10

    quasar987

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    I'm using this somewhat dead thread to ask: What does QED means? D is probably for Demonstrated, but what about Q and E?

    In french, it's CQFD, which translate into "That which had to be demonstrated."
     
  12. Oct 18, 2004 #11

    Tide

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    Close! It means "that which was to be shown" or "demonstrated."
     
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