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is x and z the same?

ok divide by x^2

after division by x^2
limit x->infinity 1/x^2-5
can you do it now?

cubics dont converge

Hmm, \(0 - 5\) if we do it individually.

I got it!

ok .

But how are we allowed to divide by \(x^2\)?

5x^3/x^2 = 5x tho ....

And how did you get \(5x^3 \div x^2 = 5\)? It's 5x.

Yeah

It's the same if we divide by \(x^3\) anyway, just that the thing becomes 1/x^3

does not matter you divided both
if you take x^2 as LCM you get original back.

no it wouldn't be same.
always divide by highest power available in the question . here it is x^2

Aha! Thank you! But I still don't get how we are allowed to divide both sides by x^3.

sorry it is x^3

\[ \lim_{x \to \infty} x^3({1 \over x^3} - 5)\] \[ x^3 \lim_{x \to \infty} -5 \]

I think that you also factor the \(x^3\) out.

Oh! Wait!

\[ \lim_{x \to \infty}x^5 \times \lim_{x \to \infty}{1 \over x^3} - 5\]

x^3*

But isn't that the indeterminate form?

why are you trying to makes sense of this? a cubic does not converge

But it is given that the solution is -5.

and i say the sky has floating cotton candy for clouds; does it make it true?

http://www.wolframalpha.com/input/?i=limit%28x+to+inf%29+%281-5x%5E3%29

Uh oh. this should pretty simply be equal to negative infinity. Logically speaking.

The original question is: \[ \lim_{z \to 0} {4z^2 + z^6 \over 1- z^3} \]

if the original is that, why would you ask this other thing as a post?

Infinity, not 0.

@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

the answer to your question is -infinity @ParthKohli

Hmm, I was confused as to why it was -5 and not -infinity :P

I mean... the question that I originally asked is supposed to be -infinity, right?

yes right

Aye aye!