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anonymous

  • 2 years ago

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

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  1. anonymous
    • 2 years ago
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    @ganeshie8 ?

  2. anonymous
    • 2 years ago
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    @mathslover ?

  3. mathslover
    • 2 years ago
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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.

  4. anonymous
    • 2 years ago
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    x=-1/h and h->0

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

  6. anonymous
    • 2 years ago
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    yes it is a cube

  7. mathslover
    • 2 years ago
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    I suggest you simplify the denominator... use the identity.

  8. anonymous
    • 2 years ago
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    that mod x will open with minus?

  9. anonymous
    • 2 years ago
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    ?

  10. ganeshie8
    • 2 years ago
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    may be divide numerator and denominator by x^3

  11. myininaya
    • 2 years ago
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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

  12. myininaya
    • 2 years ago
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    the some l'hopital could be used :)

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

  14. anonymous
    • 2 years ago
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    we have to write x=-1/h and then h->0

  15. Zarkon
    • 2 years ago
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    as \(x\to \infty\) you then get \((-1)\cdot 1=-1\)

  16. Zarkon
    • 2 years ago
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    x to -infinity you get what i have above

  17. anonymous
    • 2 years ago
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    @Zarkon how did u use the approximation in the first step?

  18. anonymous
    • 2 years ago
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    @ganeshie8 ?

  19. anonymous
    • 2 years ago
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    @mathslover ?

  20. Zarkon
    • 2 years ago
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    in my first step I got rid of the 1 since for large x the 1 really contributes almost nothing

  21. Zarkon
    • 2 years ago
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    and by large x I mean large negative (obviously)

  22. anonymous
    • 2 years ago
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    @mathslover ?

  23. myininaya
    • 2 years ago
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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.

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

  25. anonymous
    • 2 years ago
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    ya @sidsiddhartha

  26. anonymous
    • 2 years ago
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    i was thinking that only

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

  28. sidsiddhartha
    • 2 years ago
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    |dw:1402332193100:dw|

  29. sidsiddhartha
    • 2 years ago
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    do u get it ? @cody_123

  30. anonymous
    • 2 years ago
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    can u write the intermediate steps too? thanks

  31. sidsiddhartha
    • 2 years ago
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    |dw:1402332508279:dw| now just substitute x=(-1/t)

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

  33. anonymous
    • 2 years ago
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    sorry t->0

  34. anonymous
    • 2 years ago
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    @sidsiddhartha ?

  35. anonymous
    • 2 years ago
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    is it correct @sidsiddhartha ?

  36. sidsiddhartha
    • 2 years ago
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    no x=-1/t so |dw:1402332969397:dw| it will be like this

  37. anonymous
    • 2 years ago
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    ok @sidsiddhartha then ?

  38. sidsiddhartha
    • 2 years ago
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    now just try to simplify it little more

  39. anonymous
    • 2 years ago
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    \[\frac{ \frac{ -1 }{ t }*\sin t+t}{ 1+t^3 }\]

  40. anonymous
    • 2 years ago
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    @sidsiddhartha ?

  41. sidsiddhartha
    • 2 years ago
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    ya good now can do it :)

  42. anonymous
    • 2 years ago
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    w8 a second in denominator i should be 1-t^3

  43. anonymous
    • 2 years ago
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    @sidsiddhartha ?

  44. sidsiddhartha
    • 2 years ago
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    nope its all right now just use limit t tends to 0 (sint/t)=1

  45. anonymous
    • 2 years ago
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    @sidsiddhartha

  46. anonymous
    • 2 years ago
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    \[1+\frac{ -1 }{ t^3 }=\frac{ t^3-1 }{ t^3 }\]

  47. anonymous
    • 2 years ago
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    so t^3-1 will come in denominator?

  48. anonymous
    • 2 years ago
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    @sidsiddhartha ?

  49. sidsiddhartha
    • 2 years ago
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    |dw:1402334075049:dw| its not "-"

  50. anonymous
    • 2 years ago
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    oh silly mistake :P

  51. anonymous
    • 2 years ago
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    \[\frac{ \frac{ -1 }{ t }*\sin t+t }{ 1+t^2 }\]

  52. anonymous
    • 2 years ago
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    @sidsiddhartha ?

  53. sidsiddhartha
    • 2 years ago
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    it should be (t^3+1)

  54. anonymous
    • 2 years ago
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    where?

  55. anonymous
    • 2 years ago
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    ya in the denominator

  56. anonymous
    • 2 years ago
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    \[\frac{ t-1 }{ t^3+1 }\]

  57. anonymous
    • 2 years ago
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    -1

  58. sidsiddhartha
    • 2 years ago
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    yeah :)

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