Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

angela210793 Group Title

Find a and b

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1336063713485:dw|

    • 2 years ago
  2. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Hmm \[\frac{x^3 + 2x-ax^3+b + ax -bx^2}{x^2-1}\]

    • 2 years ago
  3. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    then?

    • 2 years ago
  4. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Divide it by x^2

    • 2 years ago
  5. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  6. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    x^3 - ax^3 = 0 => a= 1

    • 2 years ago
  7. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    where did u get tht???????

    • 2 years ago
  8. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    \[\large \frac{\frac{x^3 + 2x-ax^3+b + ax -bx^2}{x^2}}{\frac{x^2-1}{x^2}} = \frac{x-ax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} -b}{1 - \frac{1}{x^2}}\]

    • 2 years ago
  9. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ishaan y u divided it by x^2???

    • 2 years ago
  10. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

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

    • 2 years ago
  11. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

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

    • 2 years ago
  12. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Oh sorry, Typo. a=1. not a=0

    • 2 years ago
  13. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  14. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes. :-)

    • 2 years ago
  15. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i actually got the same from wht i did

    • 2 years ago
  16. vancouver2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ishan can you also help me with my pboelm?

    • 2 years ago
  17. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Yeah sure. Vancouver

    • 2 years ago
  18. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    but...u get infinite lim x->inf(x-ax-b)=0 how did u do it then?

    • 2 years ago
  19. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    Link Me :D

    • 2 years ago
  20. vancouver2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Iright in your left in open questions ;)

    • 2 years ago
  21. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    \[\frac{x-ax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} -b}{1 - \frac{1}{x^2}}\]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 \(\frac{-b}1\).

    • 2 years ago
  22. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    And a=1, b=0 are the solutions

    • 2 years ago
  23. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @Ishaan Y I no get it ??????

    • 2 years ago
  24. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ishaan u got a random value of a??? O.o

    • 2 years ago
  25. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    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.

    • 2 years ago
  26. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    for* not from

    • 2 years ago
  27. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ohhh.....silly me :S :$ :$ got it got it :$:$:$:$

    • 2 years ago
  28. angela210793 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you Ishaan :D

    • 2 years ago
  29. Ishaan94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 5

    You're Welcome! :D

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.