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angela210793Best ResponseYou've already chosen the best response.1
dw:1336063713485:dw
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
Hmm \[\frac{x^3 + 2xax^3+b + ax bx^2}{x^21}\]
 one year ago

angela210793Best ResponseYou've already chosen the best response.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
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
x^3  ax^3 = 0 => a= 1
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
where did u get tht???????
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
\[\large \frac{\frac{x^3 + 2xax^3+b + ax bx^2}{x^2}}{\frac{x^21}{x^2}} = \frac{xax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} b}{1  \frac{1}{x^2}}\]
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
Ishaan y u divided it by x^2???
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
\[x\to \infty\] so, if xax isn't Zero the limit would approach infinity. And xax can be Zero only when a=0.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
Hmm it's the matter of choice angela. I felt dividing it with \(x^2\) would be easier.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
Oh sorry, Typo. a=1. not a=0
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
ohhh i got it _ silly me....u did tht so u get lim of those who have x in the denom zero right?
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
i actually got the same from wht i did
 one year ago

vancouver2012Best ResponseYou've already chosen the best response.0
Ishan can you also help me with my pboelm?
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
but...u get infinite lim x>inf(xaxb)=0 how did u do it then?
 one year ago

vancouver2012Best ResponseYou've already chosen the best response.0
Iright in your left in open questions ;)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
\[\frac{xax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} b}{1  \frac{1}{x^2}}\]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 \(\frac{b}1\).
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
And a=1, b=0 are the solutions
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
@Ishaan Y I no get it ??????
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
Ishaan u got a random value of a??? O.o
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.5
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.
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
ohhh.....silly me :S :$ :$ got it got it :$:$:$:$
 one year ago

angela210793Best ResponseYou've already chosen the best response.1
Thank you Ishaan :D
 one year ago
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