## ParthKohli 4 years ago How is $\lim_{x \to \infty} 1 - 5x^3 = -5$?

1. amistre64

is x and z the same?

2. anonymous

ok divide by x^2

3. anonymous

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

4. amistre64

cubics dont converge

5. ParthKohli

Hmm, $$0 - 5$$ if we do it individually.

6. ParthKohli

I got it!

7. anonymous

ok .

8. ParthKohli

But how are we allowed to divide by $$x^2$$?

9. amistre64

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

10. ParthKohli

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

11. ParthKohli

Yeah

12. ParthKohli

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

13. anonymous

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

14. anonymous

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

15. ParthKohli

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

16. anonymous

sorry it is x^3

17. ParthKohli

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

18. ParthKohli

I think that you also factor the $$x^3$$ out.

19. ParthKohli

Oh! Wait!

20. ParthKohli

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

21. ParthKohli

x^3*

22. ParthKohli

But isn't that the indeterminate form?

23. amistre64

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

24. ParthKohli

But it is given that the solution is -5.

25. amistre64

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

26. 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.

27. amistre64
28. 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).

29. anonymous

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

30. ParthKohli

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

31. amistre64

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

32. ParthKohli

Infinity, not 0.

33. 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

34. ParthKohli

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

35. anonymous

36. ParthKohli

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

37. ParthKohli

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

38. anonymous

yes right

39. anonymous

Aye aye!