angela210793
  • angela210793
Find a and b
Mathematics
jamiebookeater
  • jamiebookeater
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angela210793
  • angela210793
|dw:1336063713485:dw|
anonymous
  • anonymous
Hmm \[\frac{x^3 + 2x-ax^3+b + ax -bx^2}{x^2-1}\]
angela210793
  • angela210793
then?

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anonymous
  • anonymous
Divide it by x^2
angela210793
  • angela210793
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
anonymous
  • anonymous
x^3 - ax^3 = 0 => a= 1
angela210793
  • angela210793
where did u get tht???????
anonymous
  • anonymous
\[\large \frac{\frac{x^3 + 2x-ax^3+b + ax -bx^2}{x^2}}{\frac{x^2-1}{x^2}} = \frac{x-ax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} -b}{1 - \frac{1}{x^2}}\]
angela210793
  • angela210793
Ishaan y u divided it by x^2???
anonymous
  • anonymous
\[x\to \infty\] so, if x-ax isn't Zero the limit would approach infinity. And x-ax can be Zero only when a=0.
anonymous
  • anonymous
Hmm it's the matter of choice angela. I felt dividing it with \(x^2\) would be easier.
anonymous
  • anonymous
Oh sorry, Typo. a=1. not a=0
angela210793
  • angela210793
ohhh i got it -_- silly me....u did tht so u get lim of those who have x in the denom zero right?
anonymous
  • anonymous
Yes. :-)
angela210793
  • angela210793
i actually got the same from wht i did
anonymous
  • anonymous
Ishan can you also help me with my pboelm?
anonymous
  • anonymous
Yeah sure. Vancouver
angela210793
  • angela210793
but...u get infinite lim x->inf(x-ax-b)=0 how did u do it then?
anonymous
  • anonymous
Link Me :D
anonymous
  • anonymous
Iright in your left in open questions ;)
anonymous
  • anonymous
\[\frac{x-ax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} -b}{1 - \frac{1}{x^2}}\]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 \(\frac{-b}1\).
anonymous
  • anonymous
And a=1, b=0 are the solutions
angela210793
  • angela210793
@Ishaan Y I no get it ??????
angela210793
  • angela210793
Ishaan u got a random value of a??? O.o
anonymous
  • anonymous
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.
anonymous
  • anonymous
for* not from
angela210793
  • angela210793
ohhh.....silly me :S :$ :$ got it got it :$:$:$:$
angela210793
  • angela210793
Thank you Ishaan :D
anonymous
  • anonymous
You're Welcome! :D

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