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cluo

  • 3 years ago

limit of 17/14^x+25arctan(x^5) as x approaches infinity.

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  1. cluo
    • 3 years ago
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    its the first one

  2. cluo
    • 3 years ago
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    17/14^x is zero as x goes to infinity?

  3. cluo
    • 3 years ago
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    then 25arctan(x^5) is left right?

  4. cluo
    • 3 years ago
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    so can i get the solution without looking at the graph?

  5. walters
    • 3 years ago
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    |dw:1360488025343:dw|

  6. agent0smith
    • 3 years ago
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    Yeah, arctan(x^5) will approach the same limit as arctan(x)

  7. agent0smith
    • 3 years ago
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    @walters i don't think that's the correct function

  8. agent0smith
    • 3 years ago
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    It is this, right??\[\frac{ 17 }{ 14^x}+25*\arctan(x^5)\]

  9. cluo
    • 3 years ago
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    yes

  10. agent0smith
    • 3 years ago
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    lim x=> inf for arctan(x) is pi/2 if i recall correctly. Since we have 25arctan(x)... So 25pi/2.

  11. walters
    • 3 years ago
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    ok it means it will be|dw:1360488400036:dw|

  12. agent0smith
    • 3 years ago
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    Hehe, still not the right function @walters

  13. walters
    • 3 years ago
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    this means the answer is infinity sice the limit of the first part is infinity which will affect the second part of the function

  14. cluo
    • 3 years ago
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    how do you know the limit of arctan is pi/2?

  15. agent0smith
    • 3 years ago
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    @walters this is the function \[\frac{ 17 }{ 14^x}+25*\arctan(x^5) \]

  16. agent0smith
    • 3 years ago
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    lim of arctanx is pi/2 because... tanx has asymptotes at pi/2, and arctanx is its inverse. There might be more proof needed than that though...

  17. cluo
    • 3 years ago
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    i can see it is pi/2 on a graph when x goes to infinity but how to find it without a graph? or do you just memorize it?

  18. agent0smith
    • 3 years ago
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    @cluo tan(x) is undefined for pi/2 radians. As x => pi/2, tanx approaches infinity. Therefore it's inverse, arctanx, is bounded between y= -pi/2 and y= pi/2

  19. walters
    • 3 years ago
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    ok i see is my mistake |dw:1360488999424:dw| thx @agent0smith

  20. agent0smith
    • 3 years ago
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    there you go :) you had the limit correct in your earlier post, 25pi/2, just had the function written incorrectly.

  21. agent0smith
    • 3 years ago
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    Make sense @cluo? The reason for the limit of arctanx as x => inf. is just due to arctanx being the inverse of tanx (tanx restricted to a domain -pi/2 to pi/2)... since tanx is not one-to-one, and thus doesn't have a valid inverse function w/o restriction.

  22. cluo
    • 3 years ago
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    kind of but not really

  23. agent0smith
    • 3 years ago
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    I'll get a graph... do you remember much on inverse functions?

  24. cluo
    • 3 years ago
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    if i look at a graph then i see it, nope everything is hazy. how do you retain that information after years?

  25. agent0smith
    • 3 years ago
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    haha, it took me a few mins to remember it, it wasn't instant. Maybe this will help: http://ocw.mit.edu/ans7870/18/18.013a/textbook/chapter01/images/tan_atan.gif See how the arctanx is *only* that part of tanx between x=-pi/2 and pi/2?

  26. agent0smith
    • 3 years ago
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    Plus it might be easier to remember that tan90 (degrees) is undefined.

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