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sami21
 2 years ago
Best ResponseYou've already chosen the best response.3after division by x^2 limit x>infinity 1/x^25 can you do it now?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm, \(0  5\) if we do it individually.

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.05x^3/x^2 = 5x tho ....

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

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

sami21
 2 years ago
Best ResponseYou've already chosen the best response.3does not matter you divided both if you take x^2 as LCM you get original back.

sami21
 2 years ago
Best ResponseYou've already chosen the best response.3no it wouldn't be same. always divide by highest power available in the question . here it is x^2

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \lim_{x \to \infty} x^3({1 \over x^3}  5)\] \[ x^3 \lim_{x \to \infty} 5 \]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I think that you also factor the \(x^3\) out.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \lim_{x \to \infty}x^5 \times \lim_{x \to \infty}{1 \over x^3}  5\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0But isn't that the indeterminate form?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0why are you trying to makes sense of this? a cubic does not converge

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0But it is given that the solution is 5.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0and i say the sky has floating cotton candy for clouds; does it make it true?

sami21
 2 years ago
Best ResponseYou've already chosen the best response.3no 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.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=limit%28x+to+inf%29+%2815x%5E3%29

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Well, 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).

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.0Uh oh. this should pretty simply be equal to negative infinity. Logically speaking.

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0if the original is that, why would you ask this other thing as a post?

sami21
 2 years ago
Best 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

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

sami21
 2 years ago
Best ResponseYou've already chosen the best response.3the answer to your question is infinity @ParthKohli

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm, I was confused as to why it was 5 and not infinity :P

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I mean... the question that I originally asked is supposed to be infinity, right?
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