How is \[ \lim_{x \to \infty} 1 - 5x^3 = -5\]?

- ParthKohli

How is \[ \lim_{x \to \infty} 1 - 5x^3 = -5\]?

- katieb

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- amistre64

is x and z the same?

- anonymous

ok divide by x^2

- 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

cubics dont converge

- ParthKohli

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

- ParthKohli

I got it!

- anonymous

ok .

- ParthKohli

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

- amistre64

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

- ParthKohli

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

- ParthKohli

Yeah

- ParthKohli

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

- anonymous

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

- anonymous

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

- ParthKohli

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

- anonymous

sorry it is x^3

- ParthKohli

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

- ParthKohli

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

- ParthKohli

Oh! Wait!

- ParthKohli

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

- ParthKohli

x^3*

- ParthKohli

But isn't that the indeterminate form?

- amistre64

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

- ParthKohli

But it is given that the solution is -5.

- amistre64

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

- 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

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

- 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

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

- ParthKohli

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

- amistre64

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

- ParthKohli

Infinity, not 0.

- 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

##### 1 Attachment

- ParthKohli

Because they showed something like \[ \lim_{z \to \infty} {4z^2 + z^6 \over 1 - z^3} = {-\infty \over -5} \]

- anonymous

the answer to your question is -infinity @ParthKohli

- ParthKohli

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

- ParthKohli

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

- anonymous

yes right

- apoorvk

Aye aye!

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