Question

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

Answer 1

is x and z the same?

Answer 2

ok divide by x^2

Answer 3

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

Answer 4

cubics dont converge

Answer 5

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

Answer 6

I got it!

Answer 7

ok .

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Yeah

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

sorry it is x^3

Answer 17

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

Answer 18

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

Answer 19

Oh! Wait!

Answer 20

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

Answer 21

x^3*

Answer 22

But isn't that the indeterminate form?

Answer 23

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

Answer 24

But it is given that the solution is -5.

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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).

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

Infinity, not 0.

Answer 33

@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

Answer 34

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

Answer 35

the answer to your question is -infinity @ParthKohli

Answer 36

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

Answer 37

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

Answer 38

yes right

Answer 39

Aye aye!