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Limit with variable exponents

  1. Mar 18, 2013 #1

    B18

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    1. The problem statement, all variables and given/known data
    Determine if sequence converges or diverges, if it converges find its limit.
    a (sub) n= (3^(n+2))/(5^(n))


    3. The attempt at a solution
    The only things Ive tried doing thus far is setting the sequence up as a function and letting x approach infinity. I then tried using l hospitals rule and this really didn't help me prove that this functions limit is equal to 0. I eventually tried putting the limit onto the numerator and denominator. This still really didn't help. Lastly I put the limit into the exponents in both numerator and denominator and get 3^∞/5^∞. Was I on the right path at any time here?

    -thanks for any suggestions!
     
  2. jcsd
  3. Mar 18, 2013 #2

    jbunniii

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    Hint:
    $$\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$$
     
  4. Mar 18, 2013 #3

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    are you hinting I apply this to the original sequence? Or (3/5)^∞.
     
  5. Mar 18, 2013 #4

    jbunniii

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    ##(3/5)^{\infty}## is meaningless. Apply it to the original sequence.
     
  6. Mar 18, 2013 #5

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    just not seeing how your applying that to the original sequence. The exponents are different. n+2, and n.
     
  7. Mar 18, 2013 #6

    jbunniii

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    Can you write ##3^{n+2}## as the product of two factors?
     
  8. Mar 18, 2013 #7

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    Alright so here is what i get.
    lim x->∞ 3^(n+2)/5^(n)
    lim x->∞ 3^(n)X3^(2)/5^(n)
    from this point I struggle to understand what the limit of 3^n, and 5^n is. I assume they are zero because the limit of the sequence is zero. But what makes them zero?
     
  9. Mar 18, 2013 #8

    jbunniii

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    Well, apply the first hint I gave you.
    $$\frac{3^{n+2}}{5^{n}} = \frac{(3^n)(3^2)}{5^n} = ???$$
     
  10. Mar 18, 2013 #9
    You have some pretty fundamental misunderstandings going on here.

    1) What was your thought process for trying L'Hopitals rule? L'hopitals rule has no meaning here.

    2) Do you know what it means for a sequence to converge or diverge? If so, write the definition, formally or not.
     
  11. Mar 18, 2013 #10

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    sorry i had got to this point.
    Its equal to 9 lim x->∞ (3/5)^x
    the limit is equal to 0 so it would be 9 times 0. How can i show that the limit of (3/5)^x is 0 as x approaches infinity?
    i understand as x gets bigger the answer gets very small but how can I prove this statement?
     
  12. Mar 18, 2013 #11

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    I thought l hospitals was applicable at first when i plugged in ∞ for n and got 3^(∞+2)/5^(∞)=∞/∞.
    Convergence and divergence is just determined by the result of the limit. If it exists the limit converges if it DNE it diverges. We are just starting this section in class I may be jumping ahead a bit by doing this problem.
     
  13. Mar 18, 2013 #12

    SammyS

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    Do you know that [itex]\displaystyle \ \ \frac{a^k}{b^k}=\left(\frac{a}{b}\right)^k\ ?[/itex]
     
  14. Mar 18, 2013 #13

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    yes I understand that exponent rule.
     
  15. Mar 18, 2013 #14

    HallsofIvy

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    The crucial point is that 3/5< 1. If you want to prove that the limit is 0 (which is NOT the problem you originally posted: "Determine if sequence converges or diverges, if it converges find its limit") the use the definition: given [itex]\epsilon> 0[/itex], there exist N such that if n> N then [itex]|9(3/5)^n|= 9(3/5)^n< \epsilon[/itex]. Can you solve that for n?
     
  16. Mar 18, 2013 #15

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    Im fairly certain this is a topic I struggled with in calc 1. I understand limits fairly well, just have trouble with exponential functions like 6^(x), 7^(x+2) ect. I don't see why these functions aren't indeterminate when x approaches infinity by 1^(x) is.
     
  17. Mar 18, 2013 #16

    jbunniii

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    It is a special case of a more general statement:
    $$\lim_{n \rightarrow \infty} a^n = 0$$
    for any real number ##a## such that ##-1 < a < 1##. Do you know this theorem, or can you prove it?
     
  18. Mar 18, 2013 #17

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    Sorry that i've created such confusion. I do understand that this sequence converges to 0. Convergence and divergence is not the issue anymore. However I am trying to understand how limits with variable exponents work. Since 3/5 is < 1 this makes the function exponential decay making it go to zero. If it were (5/3)^x this function would go to infinity because (5/3) is >1.
     
  19. Mar 18, 2013 #18

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    That was never a theorem I knew or understood till this point, I think the problem here is that im trying to search for a way to show this limit (3/5)^x as x appr. ∞=0 when there is no in-between steps to show that.
     
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