angela210793
Find a and b



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angela210793
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dw:1336063713485:dw

Ishaan94
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Hmm \[\frac{x^3 + 2xax^3+b + ax bx^2}{x^21}\]

angela210793
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then?

Ishaan94
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Divide it by x^2

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

Ishaan94
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x^3  ax^3 = 0
=> a= 1

angela210793
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where did u get tht???????

Ishaan94
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\[\large \frac{\frac{x^3 + 2xax^3+b + ax bx^2}{x^2}}{\frac{x^21}{x^2}} = \frac{xax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} b}{1  \frac{1}{x^2}}\]

angela210793
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Ishaan y u divided it by x^2???

Ishaan94
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\[x\to \infty\]
so, if xax isn't Zero the limit would approach infinity. And xax can be Zero only when a=0.

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

Ishaan94
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Oh sorry, Typo. a=1. not a=0

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?

Ishaan94
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Yes. :)

angela210793
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i actually got the same from wht i did

vancouver2012
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Ishan can you also help me with my pboelm?

Ishaan94
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Yeah sure. Vancouver

angela210793
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but...u get infinite lim x>inf(xaxb)=0 how did u do it then?

Ishaan94
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Link Me :D

vancouver2012
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Iright in your left in open questions ;)

Ishaan94
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\[\frac{xax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} b}{1  \frac{1}{x^2}}\]With a=1\[\large \frac{x1\cdot x +2\frac1{x} + \frac b{x^2} + \frac {1}{x^2} b}{1  \frac{1}{x^2}} = \frac{2\frac1x + \frac{b}{x^2} + \frac1{x^2} b}{1\frac1{x^2}}\]
\[x\to \infty \implies \frac{1}{x}\to 0\]
The limit is now \(\frac{b}1\).

Ishaan94
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And a=1, b=0 are the solutions

angela210793
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@Ishaan Y I no get it ??????

angela210793
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Ishaan u got a random value of a??? O.o

Ishaan94
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It's not a random value. \(xax\). and \(x\to \infty\). It's important form the both terms to cancel out. Which is only possible if a=1.

Ishaan94
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for* not from

angela210793
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ohhh.....silly me :S :$ :$ got it got it :$:$:$:$

angela210793
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Thank you Ishaan :D

Ishaan94
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You're Welcome! :D