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Help finding where a sequence converges

  1. Sep 30, 2012 #1
    Hi all,

    I have a general question along with a specific one. Could anyone give me an intuitive explanation of finding limits of convergent sequences? I have a test and just do not understand how to consistently find the answer to these problems. I can understand intuitively obvious ones, but for ones like the example below, I am stumped.

    Here's my specific question-

    This sequence converges at (1/2)sqrt(2). How would one find this?

    (n^2)/sqrt((2n^4)+1)

    Thanks!

    Rob
     
  2. jcsd
  3. Sep 30, 2012 #2

    Dick

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    Pull a factor of n^2 out the denominator and cancel it with the factor of n^2 in the numerator. What's left?
     
  4. Sep 30, 2012 #3
    You are a genius. Haha. Thanks!
     
  5. Oct 1, 2012 #4
    Actually, I still do not understand. Removing a term of n^2 will give 1/(2n^2+1)^(1/2).

    How would you get sqrt(2)/2 from this?
     
  6. Oct 1, 2012 #5
    Would the answer not be 1/2?

    My book may have an error.

    Sorry for the double post.
     
  7. Oct 1, 2012 #6

    Dick

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    Your algebra is wrong. To get a factor of [itex]n^2[/itex] out of [itex]\sqrt{2n^4+1}[/itex] first write it as [itex]\sqrt{n^4(2+\frac{1}{n^4})}[/itex]. Now what??
     
  8. Oct 1, 2012 #7
    The n^2 on top and bottom cancel. I'm left with 1/sqrt(2+(1/n^4)). Still not sure what to do next.

    Thanks for your patience!
     
  9. Oct 1, 2012 #8

    Dick

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    Now take the limit n->infinity. What happens to the 1/n^4?
     
  10. Oct 1, 2012 #9
    It goes to zero, and I'm left with 1/sqrt(2). Ohhhh, and that is equivalent to sqrt(2)/2, yes?
     
  11. Oct 1, 2012 #10

    Dick

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    Yes it is!
     
  12. Oct 1, 2012 #11
    HaHA! Thanks a bunch! I have a test later and will definitely utilize my newfound limit skills.
     
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