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amistre64Best ResponseYou've already chosen the best response.0
is x and z the same?
 one year ago

sami21Best ResponseYou've already chosen the best response.3
after division by x^2 limit x>infinity 1/x^25 can you do it now?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
cubics dont converge
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hmm, \(0  5\) if we do it individually.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But how are we allowed to divide by \(x^2\)?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
5x^3/x^2 = 5x tho ....
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
And how did you get \(5x^3 \div x^2 = 5\)? It's 5x.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
It's the same if we divide by \(x^3\) anyway, just that the thing becomes 1/x^3
 one year ago

sami21Best ResponseYou've already chosen the best response.3
does not matter you divided both if you take x^2 as LCM you get original back.
 one year ago

sami21Best ResponseYou've already chosen the best response.3
no it wouldn't be same. always divide by highest power available in the question . here it is x^2
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Aha! Thank you! But I still don't get how we are allowed to divide both sides by x^3.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[ \lim_{x \to \infty} x^3({1 \over x^3}  5)\] \[ x^3 \lim_{x \to \infty} 5 \]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I think that you also factor the \(x^3\) out.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[ \lim_{x \to \infty}x^5 \times \lim_{x \to \infty}{1 \over x^3}  5\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But isn't that the indeterminate form?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
why are you trying to makes sense of this? a cubic does not converge
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But it is given that the solution is 5.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
and i say the sky has floating cotton candy for clouds; does it make it true?
 one year ago

sami21Best ResponseYou've already chosen the best response.3
no you cannot take it out always.just divide each term because if you take common then you will get x^3also then it will always gives you infinity.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=limit%28x+to+inf%29+%2815x%5E3%29
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Well, may I know how you are allowed to divide by \(x^3\)? (I know that you divide the variable with the exponent as the highest power).
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
Uh oh. this should pretty simply be equal to negative infinity. Logically speaking.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
The original question is: \[ \lim_{z \to 0} {4z^2 + z^6 \over 1 z^3} \]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
if the original is that, why would you ask this other thing as a post?
 one year ago

sami21Best ResponseYou've already chosen the best response.3
@amistre64 you are right .it does not converges . this division rule works only when you have fractions .rational functions @ParthKohli you should take common x^3. for example
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Because they showed something like \[ \lim_{z \to \infty} {4z^2 + z^6 \over 1  z^3} = {\infty \over 5} \]
 one year ago

sami21Best ResponseYou've already chosen the best response.3
the answer to your question is infinity @ParthKohli
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hmm, I was confused as to why it was 5 and not infinity :P
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I mean... the question that I originally asked is supposed to be infinity, right?
 one year ago
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