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Help with using the Squeeze Theorem to find a limit.

  1. Oct 27, 2008 #1
    We were given the hint to try using the Squeeze Theorem in order to find the
    limit as n → ∞ of the sequence {(2^n + 3^n)^(1/n)}.

    I understand the concept of the squeeze theorem that I need to find functions A and B such that A ≤ {(2^n + 3^n)^(1/n)} ≤ B, and A and B limit to the same quantity, say "L."
    So lim A = lim B = L, so that I can conclude that lim (2^n + 3^n)^(1/n) = L.

    I don't know how to come up with those functions. It has been 2 years since I last took a calculus class, so I am very rusty with limits.

    So far all I have is that 0 ≤ {(2^n + 3^n)^(1/n)} ≤ 2^n + 3^n.

    So I can say 0 limits to 0, but then how would I evaluate
    the limit of 2^n + 3^n as n approaches ∞? It would just keep getting bigger so I would have that limit as ∞. So I am stuck with 0 and ∞ as limits which is wrong because A and B are supposed to limit to the same value.

    Any help, tips, corrections, and/or suggestions is greatly appreciated.
    Thank you for your time!
     
  2. jcsd
  3. Oct 27, 2008 #2

    gabbagabbahey

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    To find A and B, you might try using the Binomial theorem on [itex](2^n+3^n)^{\frac{1}{n}}=3\left(1+\left(\frac{2}{3}\right)^n \right)^{\frac{1}{n}}[/itex]...Clearly, [itex]\left(1+\left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}} \geq 1[/itex] can you think of a function that is always greater than or equal to this quantity and also has a limit of 1?
     
  4. Oct 27, 2008 #3

    Dick

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    I'm not sure I'd use the squeeze theorem on this. But if you want to, 3^n<=2^n+3^n<=3^n+3^n.
     
  5. Oct 27, 2008 #4

    Dick

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    I think the binomial theorem is just plain overkill. If you want to do it that way, then just take the log of the second factor and look at the limit. BTW that's how I would have done it without using 'squeeze'.
     
    Last edited: Oct 27, 2008
  6. Oct 28, 2008 #5
    Thanks for looking at this thread. We haven't gotten to the Binomial Theorem yet, and I don't think our professor will be emphasizing it that much, so I think that if I use that theorem, it would appear unrealistic for me to do that since we haven't covered it yet.

    Thanks for the input though, and I'll definitely look into the binomial theorem a bit more when our professor mentions it.
     
  7. Oct 28, 2008 #6
    Will this work?
    3 [tex]\leq\sqrt[n]{2^{n}+3^{n}}[/tex] [tex]\leq[/tex] 3[tex]\sqrt[n]{2}[/tex]

    How do I know the limit of 3[tex]\sqrt[n]{2}[/tex] is 3?

    So is the limit of [tex]3^{n}[/tex] = 3 = limit of [tex]3^{n}[/tex] + [tex]3^{n}[/tex].
    But I am taking [tex]\sqrt[n]{2^{n}+3^{n}}[/tex] and not just [tex]{2^{n}+3^{n}}[/tex]
     
  8. Oct 28, 2008 #7

    Dick

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    To figure out the lim (2)^(1/n), take the log. The limit if the log is pretty clear, isn't it?
     
  9. Oct 28, 2008 #8
    I've forgotten how to take logs and limits of logs.
    Is it
    log 2^(1/n) can be rewritten as (1/n) = [tex]log_{2}[/tex] x?
    How do I find the limit from there? Take ln?
     
  10. Oct 28, 2008 #9

    Dick

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    In (1/n)=log_2(x) do you mean x=2^(1/n)???
     
    Last edited: Oct 28, 2008
  11. Oct 28, 2008 #10

    Dick

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    If you want to try and sneak around the logs, 2^n+3^n<=3^n+3^n+3^n=3*3^n=3^(n+1).
     
  12. Oct 28, 2008 #11
    Where's the radical?

    I'm still not quite "seeing" it. Where does the radical 'nth root' go?
    I understand the comparison of 2^n+3^n<=3^n+3^n+3^n=3*3^n=3^(n+1), but I'm still struggling with how to relate it to the problem that has [tex](2^{n} + 3^{n})^{1/n}[/tex]. Where are you putting the (1/n) exponent in the comparison?
     
  13. Oct 28, 2008 #12

    Dick

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    (3^n)^(1/n)=3<=(2^n+3^n)^(1/n)<=(3^(n+1))^(1/n)=3^(1+1/n).
     
  14. Oct 28, 2008 #13

    Cool, thanks a lot! I think I got it now. If not, I'll post again here. Thanks for all your help thus far!
     
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