ParthKohli
How is \[ \lim_{x \to \infty} 1 - 5x^3 = -5\]?
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amistre64
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is x and z the same?
sami-21
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ok divide by x^2
sami-21
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after division by x^2
limit x->infinity 1/x^2-5
can you do it now?
amistre64
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cubics dont converge
ParthKohli
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Hmm, \(0 - 5\) if we do it individually.
ParthKohli
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I got it!
sami-21
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ok .
ParthKohli
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But how are we allowed to divide by \(x^2\)?
amistre64
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5x^3/x^2 = 5x tho ....
ParthKohli
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And how did you get \(5x^3 \div x^2 = 5\)? It's 5x.
ParthKohli
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Yeah
ParthKohli
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It's the same if we divide by \(x^3\) anyway, just that the thing becomes 1/x^3
sami-21
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does not matter you divided both
if you take x^2 as LCM you get original back.
sami-21
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no it wouldn't be same.
always divide by highest power available in the question . here it is x^2
ParthKohli
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Aha! Thank you! But I still don't get how we are allowed to divide both sides by x^3.
sami-21
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sorry it is x^3
ParthKohli
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\[ \lim_{x \to \infty} x^3({1 \over x^3} - 5)\] \[ x^3 \lim_{x \to \infty} -5 \]
ParthKohli
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I think that you also factor the \(x^3\) out.
ParthKohli
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Oh! Wait!
ParthKohli
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\[ \lim_{x \to \infty}x^5 \times \lim_{x \to \infty}{1 \over x^3} - 5\]
ParthKohli
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x^3*
ParthKohli
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But isn't that the indeterminate form?
amistre64
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why are you trying to makes sense of this? a cubic does not converge
ParthKohli
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But it is given that the solution is -5.
amistre64
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and i say the sky has floating cotton candy for clouds; does it make it true?
sami-21
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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.
ParthKohli
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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
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Uh oh. this should pretty simply be equal to negative infinity. Logically speaking.
ParthKohli
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The original question is: \[ \lim_{z \to 0} {4z^2 + z^6 \over 1- z^3} \]
amistre64
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if the original is that, why would you ask this other thing as a post?
ParthKohli
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Infinity, not 0.
sami-21
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@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
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Because they showed something like \[ \lim_{z \to \infty} {4z^2 + z^6 \over 1 - z^3} = {-\infty \over -5} \]
sami-21
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the answer to your question is -infinity @ParthKohli
ParthKohli
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Hmm, I was confused as to why it was -5 and not -infinity :P
ParthKohli
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I mean... the question that I originally asked is supposed to be -infinity, right?
sami-21
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yes right
apoorvk
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Aye aye!