## ParthKohli Group Title How is $\lim_{x \to \infty} 1 - 5x^3 = -5$? 2 years ago 2 years ago

1. amistre64 Group Title

is x and z the same?

2. sami-21 Group Title

ok divide by x^2

3. sami-21 Group Title

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

4. amistre64 Group Title

cubics dont converge

5. ParthKohli Group Title

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

6. ParthKohli Group Title

I got it!

7. sami-21 Group Title

ok .

8. ParthKohli Group Title

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

9. amistre64 Group Title

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

10. ParthKohli Group Title

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

11. ParthKohli Group Title

Yeah

12. ParthKohli Group Title

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

13. sami-21 Group Title

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

14. sami-21 Group Title

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

15. ParthKohli Group Title

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

16. sami-21 Group Title

sorry it is x^3

17. ParthKohli Group Title

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

18. ParthKohli Group Title

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

19. ParthKohli Group Title

Oh! Wait!

20. ParthKohli Group Title

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

21. ParthKohli Group Title

x^3*

22. ParthKohli Group Title

But isn't that the indeterminate form?

23. amistre64 Group Title

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

24. ParthKohli Group Title

But it is given that the solution is -5.

25. amistre64 Group Title

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

26. sami-21 Group Title

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 Group Title
28. ParthKohli Group Title

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. apoorvk Group Title

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

30. ParthKohli Group Title

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

31. amistre64 Group Title

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

32. ParthKohli Group Title

Infinity, not 0.

33. sami-21 Group Title

@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 Group Title

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

35. sami-21 Group Title

36. ParthKohli Group Title

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

37. ParthKohli Group Title

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

38. sami-21 Group Title

yes right

39. apoorvk Group Title

Aye aye!