ParthKohli
  • ParthKohli
How is \[ \lim_{x \to \infty} 1 - 5x^3 = -5\]?
Mathematics
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
is x and z the same?
anonymous
  • anonymous
ok divide by x^2
anonymous
  • anonymous
after division by x^2 limit x->infinity 1/x^2-5 can you do it now?

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More answers

amistre64
  • amistre64
cubics dont converge
ParthKohli
  • ParthKohli
Hmm, \(0 - 5\) if we do it individually.
ParthKohli
  • ParthKohli
I got it!
anonymous
  • anonymous
ok .
ParthKohli
  • ParthKohli
But how are we allowed to divide by \(x^2\)?
amistre64
  • amistre64
5x^3/x^2 = 5x tho ....
ParthKohli
  • ParthKohli
And how did you get \(5x^3 \div x^2 = 5\)? It's 5x.
ParthKohli
  • ParthKohli
Yeah
ParthKohli
  • ParthKohli
It's the same if we divide by \(x^3\) anyway, just that the thing becomes 1/x^3
anonymous
  • anonymous
does not matter you divided both if you take x^2 as LCM you get original back.
anonymous
  • anonymous
no it wouldn't be same. always divide by highest power available in the question . here it is x^2
ParthKohli
  • ParthKohli
Aha! Thank you! But I still don't get how we are allowed to divide both sides by x^3.
anonymous
  • anonymous
sorry it is x^3
ParthKohli
  • ParthKohli
\[ \lim_{x \to \infty} x^3({1 \over x^3} - 5)\] \[ x^3 \lim_{x \to \infty} -5 \]
ParthKohli
  • ParthKohli
I think that you also factor the \(x^3\) out.
ParthKohli
  • ParthKohli
Oh! Wait!
ParthKohli
  • ParthKohli
\[ \lim_{x \to \infty}x^5 \times \lim_{x \to \infty}{1 \over x^3} - 5\]
ParthKohli
  • ParthKohli
x^3*
ParthKohli
  • ParthKohli
But isn't that the indeterminate form?
amistre64
  • amistre64
why are you trying to makes sense of this? a cubic does not converge
ParthKohli
  • ParthKohli
But it is given that the solution is -5.
amistre64
  • amistre64
and i say the sky has floating cotton candy for clouds; does it make it true?
anonymous
  • anonymous
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.
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=limit%28x+to+inf%29+%281-5x%5E3%29
ParthKohli
  • ParthKohli
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).
apoorvk
  • apoorvk
Uh oh. this should pretty simply be equal to negative infinity. Logically speaking.
ParthKohli
  • ParthKohli
The original question is: \[ \lim_{z \to 0} {4z^2 + z^6 \over 1- z^3} \]
amistre64
  • amistre64
if the original is that, why would you ask this other thing as a post?
ParthKohli
  • ParthKohli
Infinity, not 0.
anonymous
  • anonymous
@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
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ParthKohli
  • ParthKohli
Because they showed something like \[ \lim_{z \to \infty} {4z^2 + z^6 \over 1 - z^3} = {-\infty \over -5} \]
anonymous
  • anonymous
the answer to your question is -infinity @ParthKohli
ParthKohli
  • ParthKohli
Hmm, I was confused as to why it was -5 and not -infinity :P
ParthKohli
  • ParthKohli
I mean... the question that I originally asked is supposed to be -infinity, right?
anonymous
  • anonymous
yes right
apoorvk
  • apoorvk
Aye aye!

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