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angela210793

Find a and b

  • one year ago
  • one year ago

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  1. angela210793
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    |dw:1336063713485:dw|

    • one year ago
  2. Ishaan94
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    Hmm \[\frac{x^3 + 2x-ax^3+b + ax -bx^2}{x^2-1}\]

    • one year ago
  3. angela210793
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    then?

    • one year ago
  4. Ishaan94
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    Divide it by x^2

    • one year ago
  5. angela210793
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    all i know is tht we have fractions like this and limit goes to infinite the limit will b calculated for the monomials in the highest power

    • one year ago
  6. Ishaan94
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    x^3 - ax^3 = 0 => a= 1

    • one year ago
  7. angela210793
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    where did u get tht???????

    • one year ago
  8. Ishaan94
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    \[\large \frac{\frac{x^3 + 2x-ax^3+b + ax -bx^2}{x^2}}{\frac{x^2-1}{x^2}} = \frac{x-ax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} -b}{1 - \frac{1}{x^2}}\]

    • one year ago
  9. angela210793
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    Ishaan y u divided it by x^2???

    • one year ago
  10. Ishaan94
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    \[x\to \infty\] so, if x-ax isn't Zero the limit would approach infinity. And x-ax can be Zero only when a=0.

    • one year ago
  11. Ishaan94
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    Hmm it's the matter of choice angela. I felt dividing it with \(x^2\) would be easier.

    • one year ago
  12. Ishaan94
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    Oh sorry, Typo. a=1. not a=0

    • one year ago
  13. angela210793
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    ohhh i got it -_- silly me....u did tht so u get lim of those who have x in the denom zero right?

    • one year ago
  14. Ishaan94
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    Yes. :-)

    • one year ago
  15. angela210793
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    i actually got the same from wht i did

    • one year ago
  16. vancouver2012
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    Ishan can you also help me with my pboelm?

    • one year ago
  17. Ishaan94
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    Yeah sure. Vancouver

    • one year ago
  18. angela210793
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    but...u get infinite lim x->inf(x-ax-b)=0 how did u do it then?

    • one year ago
  19. Ishaan94
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    Link Me :D

    • one year ago
  20. vancouver2012
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    Iright in your left in open questions ;)

    • one year ago
  21. Ishaan94
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    \[\frac{x-ax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} -b}{1 - \frac{1}{x^2}}\]With a=1\[\large \frac{x-1\cdot x +2\frac1{x} + \frac b{x^2} + \frac {1}{x^2} -b}{1 - \frac{1}{x^2}} = \frac{2\frac1x + \frac{b}{x^2} + \frac1{x^2} -b}{1-\frac1{x^2}}\] \[x\to \infty \implies \frac{1}{x}\to 0\] The limit is now \(\frac{-b}1\).

    • one year ago
  22. Ishaan94
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    And a=1, b=0 are the solutions

    • one year ago
  23. angela210793
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    @Ishaan Y I no get it ??????

    • one year ago
  24. angela210793
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    Ishaan u got a random value of a??? O.o

    • one year ago
  25. Ishaan94
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    It's not a random value. \(x-ax\). and \(x\to \infty\). It's important form the both terms to cancel out. Which is only possible if a=1.

    • one year ago
  26. Ishaan94
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    for* not from

    • one year ago
  27. angela210793
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    ohhh.....silly me :S :$ :$ got it got it :$:$:$:$

    • one year ago
  28. angela210793
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    Thank you Ishaan :D

    • one year ago
  29. Ishaan94
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    You're Welcome! :D

    • one year ago
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