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angela210793

  • 2 years ago

Find a and b

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  1. angela210793
    • 2 years ago
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    |dw:1336063713485:dw|

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

  3. angela210793
    • 2 years ago
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    then?

  4. Ishaan94
    • 2 years ago
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    Divide it by x^2

  5. angela210793
    • 2 years ago
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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

  6. Ishaan94
    • 2 years ago
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    x^3 - ax^3 = 0 => a= 1

  7. angela210793
    • 2 years ago
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    where did u get tht???????

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

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

  10. Ishaan94
    • 2 years ago
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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.

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

  12. Ishaan94
    • 2 years ago
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    Oh sorry, Typo. a=1. not a=0

  13. angela210793
    • 2 years ago
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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?

  14. Ishaan94
    • 2 years ago
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    Yes. :-)

  15. angela210793
    • 2 years ago
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    i actually got the same from wht i did

  16. vancouver2012
    • 2 years ago
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    Ishan can you also help me with my pboelm?

  17. Ishaan94
    • 2 years ago
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    Yeah sure. Vancouver

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

  19. Ishaan94
    • 2 years ago
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    Link Me :D

  20. vancouver2012
    • 2 years ago
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    Iright in your left in open questions ;)

  21. Ishaan94
    • 2 years ago
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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 \(\frac{-b}1\).

  22. Ishaan94
    • 2 years ago
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    And a=1, b=0 are the solutions

  23. angela210793
    • 2 years ago
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    @Ishaan Y I no get it ??????

  24. angela210793
    • 2 years ago
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    Ishaan u got a random value of a??? O.o

  25. Ishaan94
    • 2 years ago
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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.

  26. Ishaan94
    • 2 years ago
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    for* not from

  27. angela210793
    • 2 years ago
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    ohhh.....silly me :S :$ :$ got it got it :$:$:$:$

  28. angela210793
    • 2 years ago
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    Thank you Ishaan :D

  29. Ishaan94
    • 2 years ago
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    You're Welcome! :D

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