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Help finding where a sequence converges

  • Thread starter rocapp
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  • #1
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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
 

Answers and Replies

  • #2
Dick
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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
Pull a factor of n^2 out the denominator and cancel it with the factor of n^2 in the numerator. What's left?
 
  • #3
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You are a genius. Haha. Thanks!
 
  • #4
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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?
 
  • #5
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Would the answer not be 1/2?

My book may have an error.

Sorry for the double post.
 
  • #6
Dick
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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?
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??
 
  • #7
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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!
 
  • #8
Dick
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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!
Now take the limit n->infinity. What happens to the 1/n^4?
 
  • #9
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It goes to zero, and I'm left with 1/sqrt(2). Ohhhh, and that is equivalent to sqrt(2)/2, yes?
 
  • #10
Dick
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It goes to zero, and I'm left with 1/sqrt(2). Ohhhh, and that is equivalent to sqrt(2)/2, yes?
Yes it is!
 
  • #11
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HaHA! Thanks a bunch! I have a test later and will definitely utilize my newfound limit skills.
 

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