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cody_123 Group Title

\[\lim_{x \rightarrow -\infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\]

  • 3 months ago
  • 3 months ago

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  1. cody_123 Group Title
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    @ganeshie8 ?

    • 3 months ago
  2. cody_123 Group Title
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    @mathslover ?

    • 3 months ago
  3. mathslover Group Title
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    Yep.. working on it. Did you try using (a^3 + b^3) identity? not sure whether that will work or not.

    • 3 months ago
  4. cody_123 Group Title
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    x=-1/h and h->0

    • 3 months ago
  5. myininaya Group Title
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    on that |x| is that a cube? I can't read for some reason. I can click on the code.

    • 3 months ago
  6. cody_123 Group Title
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    yes it is a cube

    • 3 months ago
  7. mathslover Group Title
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    I suggest you simplify the denominator... use the identity.

    • 3 months ago
  8. cody_123 Group Title
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    that mod x will open with minus?

    • 3 months ago
  9. cody_123 Group Title
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    ?

    • 3 months ago
  10. ganeshie8 Group Title
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    may be divide numerator and denominator by x^3

    • 3 months ago
  11. myininaya Group Title
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    so since x<0 then |x|=-x and since we have |x|^3 then we could replace |x|^3 with (-x)^3=-x^3 now recall that if u->0 then sin(u)/u->1 see if you can use that here divide both top and bottom by 1/x

    • 3 months ago
  12. myininaya Group Title
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    the some l'hopital could be used :)

    • 3 months ago
  13. Zarkon Group Title
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    for large x \[\Large \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\approx \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ \left| x \right|^3}\] \[\Large=\frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ x^2\left| x \right|}\] \[\Large=\frac{ x^2\sin \frac{ 1 }{ x }+1 }{ \left| x \right|}\] \[\Large\approx\frac{ x^2\sin \frac{ 1 }{ x } }{ \left| x \right|}=\frac{x}{|x|}x\sin\frac{1}{x}\]

    • 3 months ago
  14. cody_123 Group Title
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    we have to write x=-1/h and then h->0

    • 3 months ago
  15. Zarkon Group Title
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    as \(x\to \infty\) you then get \((-1)\cdot 1=-1\)

    • 3 months ago
  16. Zarkon Group Title
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    x to -infinity you get what i have above

    • 3 months ago
  17. cody_123 Group Title
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    @Zarkon how did u use the approximation in the first step?

    • 3 months ago
  18. cody_123 Group Title
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    @ganeshie8 ?

    • 3 months ago
  19. cody_123 Group Title
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    @mathslover ?

    • 3 months ago
  20. Zarkon Group Title
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    in my first step I got rid of the 1 since for large x the 1 really contributes almost nothing

    • 3 months ago
  21. Zarkon Group Title
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    and by large x I mean large negative (obviously)

    • 3 months ago
  22. cody_123 Group Title
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    @mathslover ?

    • 3 months ago
  23. myininaya Group Title
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    I would have done it like this: (but this is because i'm not commander data like zarkon (who knows all because he is a superior being to a human) \[\lim_{x \rightarrow - \infty}\frac{\frac{x^4 \sin(\frac{1}{x})}{\frac{1}{x}}+\frac{x^2}{\frac{1}{x}}}{\frac{1}{\frac{1}{x}}-\frac{x^3}{\frac{1}{x}}}\] then use the fact that if u->0 then sin(u)/u->1 it should be pretty easy after this point though.

    • 3 months ago
  24. sidsiddhartha Group Title
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    as xtends to - infinity mod(x)=-x now try to put \[x=\frac{ -1 }{ t}\]

    • 3 months ago
  25. cody_123 Group Title
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    ya @sidsiddhartha

    • 3 months ago
  26. cody_123 Group Title
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    i was thinking that only

    • 3 months ago
  27. myininaya Group Title
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    or i guess you could have just said \[\lim_{x \rightarrow -\infty}\frac{x^4 \sin(\frac{1}{x})+x^2}{1-x^3}=\lim_{x \rightarrow -\infty}\frac{x^3 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}+x^2}{1-x^3}\]

    • 3 months ago
  28. sidsiddhartha Group Title
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    |dw:1402332193100:dw|

    • 3 months ago
  29. sidsiddhartha Group Title
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    do u get it ? @cody_123

    • 3 months ago
  30. cody_123 Group Title
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    can u write the intermediate steps too? thanks

    • 3 months ago
  31. sidsiddhartha Group Title
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    |dw:1402332508279:dw| now just substitute x=(-1/t)

    • 3 months ago
  32. cody_123 Group Title
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    \[\lim_{x \rightarrow -\infty} \frac{ (\frac{ -1 }{ t })^4*\sin(-t)+(\frac{ -1 }{ t^2 }) }{1+ \frac{-1 }{ t^3 } }\]

    • 3 months ago
  33. cody_123 Group Title
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    sorry t->0

    • 3 months ago
  34. cody_123 Group Title
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    @sidsiddhartha ?

    • 3 months ago
  35. cody_123 Group Title
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    is it correct @sidsiddhartha ?

    • 3 months ago
  36. sidsiddhartha Group Title
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    no x=-1/t so |dw:1402332969397:dw| it will be like this

    • 3 months ago
  37. cody_123 Group Title
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    ok @sidsiddhartha then ?

    • 3 months ago
  38. sidsiddhartha Group Title
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    now just try to simplify it little more

    • 3 months ago
  39. cody_123 Group Title
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    \[\frac{ \frac{ -1 }{ t }*\sin t+t}{ 1+t^3 }\]

    • 3 months ago
  40. cody_123 Group Title
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    @sidsiddhartha ?

    • 3 months ago
  41. sidsiddhartha Group Title
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    ya good now can do it :)

    • 3 months ago
  42. cody_123 Group Title
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    w8 a second in denominator i should be 1-t^3

    • 3 months ago
  43. cody_123 Group Title
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    @sidsiddhartha ?

    • 3 months ago
  44. sidsiddhartha Group Title
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    nope its all right now just use limit t tends to 0 (sint/t)=1

    • 3 months ago
  45. cody_123 Group Title
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    @sidsiddhartha

    • 3 months ago
  46. cody_123 Group Title
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    \[1+\frac{ -1 }{ t^3 }=\frac{ t^3-1 }{ t^3 }\]

    • 3 months ago
  47. cody_123 Group Title
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    so t^3-1 will come in denominator?

    • 3 months ago
  48. cody_123 Group Title
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    @sidsiddhartha ?

    • 3 months ago
  49. sidsiddhartha Group Title
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    |dw:1402334075049:dw| its not "-"

    • 3 months ago
  50. cody_123 Group Title
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    oh silly mistake :P

    • 3 months ago
  51. cody_123 Group Title
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    \[\frac{ \frac{ -1 }{ t }*\sin t+t }{ 1+t^2 }\]

    • 3 months ago
  52. cody_123 Group Title
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    @sidsiddhartha ?

    • 3 months ago
  53. sidsiddhartha Group Title
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    it should be (t^3+1)

    • 3 months ago
  54. cody_123 Group Title
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    where?

    • 3 months ago
  55. cody_123 Group Title
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    ya in the denominator

    • 3 months ago
  56. cody_123 Group Title
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    \[\frac{ t-1 }{ t^3+1 }\]

    • 3 months ago
  57. cody_123 Group Title
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    -1

    • 3 months ago
  58. sidsiddhartha Group Title
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    yeah :)

    • 3 months ago
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