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tjutta Group Title

lim x>infinite 1/x^2 S(integral sign) tan^-1 dt

  • 2 years ago
  • 2 years ago

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  1. ash2326 Group Title
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    \[\lim_{x\to \infty} \frac 1 {x^2} \int \tan^{-1} t\ dt\]Is this your question?

    • 2 years ago
  2. tjutta Group Title
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    yes sorry Im new to this :-)

    • 2 years ago
  3. ash2326 Group Title
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    No need to be sorry :)

    • 2 years ago
  4. ash2326 Group Title
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    The integral doesn't involve x, so it can be taken out. \[\int \tan^{-1} t\ dt\times \lim_{x\to \infty} \frac 1 {x^2} \] tell me what's \[\frac 1 \infty=???\]

    • 2 years ago
  5. ash2326 Group Title
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    @tjutta ???

    • 2 years ago
  6. tjutta Group Title
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    \[\infty\]

    • 2 years ago
  7. ash2326 Group Title
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    nope \[\frac 1 0= \infty\] so \[\frac 1 \infty=???\]

    • 2 years ago
  8. tjutta Group Title
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    0

    • 2 years ago
  9. ash2326 Group Title
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    good :D \[\int \tan^{-1} t\times 0=???\]

    • 2 years ago
  10. tjutta Group Title
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    SIN/COS ? :-)

    • 2 years ago
  11. ash2326 Group Title
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    if you multiply 0 by tan t then what you'd get?

    • 2 years ago
  12. tjutta Group Title
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    0

    • 2 years ago
  13. ash2326 Group Title
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    so it's 0

    • 2 years ago
  14. tjutta Group Title
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    what ? the answer is 0 :-) Ok it makes since if you put it that way but for me whos math retarded its difficult:-)

    • 2 years ago
  15. ash2326 Group Title
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    You know the integral is a variable of t, if you integrate it and multiply by zero. You'd get zero. So it's not necessary to integrate. It's 0

    • 2 years ago
  16. ash2326 Group Title
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    I doubt that this is so easy. I think you have missed some portion of the question. Maybe limits on the integral. Could you check the question again?

    • 2 years ago
  17. tjutta Group Title
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    its just as you put it so I think this is right but I have one more :-) its \[\int\limits_{}^{} \frac{ (x^3)\ }{(x-1)^2(x+2) }\] Can you help me with that?

    • 2 years ago
  18. ash2326 Group Title
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    okay, did you try this?

    • 2 years ago
  19. tjutta Group Title
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    I dont know what to do :-) should I start by divide ?

    • 2 years ago
  20. ash2326 Group Title
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    What's the expansion of (x-1)^2?

    • 2 years ago
  21. tjutta Group Title
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    (x-1)(x+1) ?

    • 2 years ago
  22. ash2326 Group Title
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    What would you get if you multiply these?

    • 2 years ago
  23. tjutta Group Title
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    ok I only have 2 ?

    • 2 years ago
  24. ash2326 Group Title
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    \[(x-1)\times (x-1)=???\]

    • 2 years ago
  25. tjutta Group Title
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    x^2 - 2x + 1

    • 2 years ago
  26. ash2326 Group Title
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    Good:) Numerator we have x^2, so add and subtract (-2x+1) Could you rewrite the integral using this?

    • 2 years ago
  27. tjutta Group Title
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    \[\int\limits_{}^{} \frac{ x }{-2x+1 } \] Im I on the right way? I know Im not very good at this at all :-)

    • 2 years ago
  28. ash2326 Group Title
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    I'm sorry, i thought it's x^2 in numerator Sorry. Let me think about this

    • 2 years ago
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