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angela210793 Group TitleBest ResponseYou've already chosen the best response.1
dw:1336063713485:dw
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
Hmm \[\frac{x^3 + 2xax^3+b + ax bx^2}{x^21}\]
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
Divide it by x^2
 2 years ago

angela210793 Group TitleBest 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
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
x^3  ax^3 = 0 => a= 1
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
where did u get tht???????
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
\[\large \frac{\frac{x^3 + 2xax^3+b + ax bx^2}{x^2}}{\frac{x^21}{x^2}} = \frac{xax +2\frac1{x} + \frac b{x^2} + \frac {a}{x^2} b}{1  \frac{1}{x^2}}\]
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
Ishaan y u divided it by x^2???
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
Oh sorry, Typo. a=1. not a=0
 2 years ago

angela210793 Group TitleBest 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?
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
i actually got the same from wht i did
 2 years ago

vancouver2012 Group TitleBest ResponseYou've already chosen the best response.0
Ishan can you also help me with my pboelm?
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
Yeah sure. Vancouver
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
but...u get infinite lim x>inf(xaxb)=0 how did u do it then?
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
Link Me :D
 2 years ago

vancouver2012 Group TitleBest ResponseYou've already chosen the best response.0
Iright in your left in open questions ;)
 2 years ago

Ishaan94 Group TitleBest 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 \(\frac{b}1\).
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
And a=1, b=0 are the solutions
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
@Ishaan Y I no get it ??????
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
Ishaan u got a random value of a??? O.o
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
for* not from
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
ohhh.....silly me :S :$ :$ got it got it :$:$:$:$
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.1
Thank you Ishaan :D
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.5
You're Welcome! :D
 2 years ago
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