A community for students.
Here's the question you clicked on:
 0 viewing
angela210793
 3 years ago
Find a and b
angela210793
 3 years ago
Find a and b

This Question is Closed

angela210793
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1336063713485:dw

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

angela210793
 3 years ago
Best ResponseYou've already chosen the best response.1all 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
 3 years ago
Best ResponseYou've already chosen the best response.5x^3  ax^3 = 0 => a= 1

angela210793
 3 years ago
Best ResponseYou've already chosen the best response.1where did u get tht???????

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

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

Ishaan94
 3 years ago
Best 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.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.5Hmm it's the matter of choice angela. I felt dividing it with \(x^2\) would be easier.

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

angela210793
 3 years ago
Best ResponseYou've already chosen the best response.1ohhh i got it _ silly me....u did tht so u get lim of those who have x in the denom zero right?

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

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

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

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

Ishaan94
 3 years ago
Best 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 \(\frac{b}1\).

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

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

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

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.5It'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.

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

angela210793
 3 years ago
Best ResponseYou've already chosen the best response.1Thank you Ishaan :D
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.