Find a and b

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

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Other answers:

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

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