## probss one year ago LIMIT! using L'Hopital's rules, evaluate lim x appraoches infinity (cos(x/2))^x^2

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1. myininaya

Do you know you can write y as e^ln(y) let me know if that helps brb

2. myininaya

so what I'm say is you can write $(\cos(x/2))^{x^2}=e^{\ln((\cos(x/2))^{x^2})}=e^{x^2 \ln(\cos(x/2))}=e^{\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}}$

3. probss

i think i forgot about it

4. myininaya

now you did to find this limit $\lim_{x \rightarrow \infty}\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}$

5. probss

wow, seems it is so complicated.

6. probss

so, we can just take the value of power of exponential? what say you?

7. probss

@myininaya

8. myininaya

Instead of writing $x^2 \ln(\cos(x/2)) \text{ as } \frac{\ln(\cos(x/2))}{\frac{1}{x^2}} \text{ it might be more preferable } \\ \text{ if we write is as } \frac{x^2}{\frac{1}{\ln(\cos(x/2))}}$ still don't know if we should use l'hospital even though it says and that is why I attempted to set the problem up in this way the reason I say that is because the limit as x approaches inf of ln(cos(x/2)) dne since lim x approaches inf of cos(x/2) does not exist since it will oscillate between -1 and 1 but pretending we can use l'hospital since it say use l'hospital i guess we will force the l'hospital

9. myininaya

@zepdrix the reason i ask is because i only see it used for cases like inf/inf or 0/0

10. myininaya

not 0/dne or dne/inf

11. zepdrix

Oh boy D: Good question.. hmm So we have a problem since ln(cosx) is undefined over and over as x->infinity, yah?

12. myininaya

yah

13. zepdrix

Grr I dunno >.< maybe lego man knows

14. myininaya

maybe sith knows i seen him around a lot doing math like it is nothing

15. SithsAndGiggles

Wolfram seems to agree; the limit doesn't exist. That's my guess

16. sweetburger

LHops international house of limits

17. KingGeorge

Not only does wolfram not say anything, but Mathematica comes up blank as well.

18. myininaya

Does that mean Mathematica doesn't have the programming to find the limit or that the limit does not exist?

19. myininaya

i believe the limit does not exist and i also believe we don't need l'hopital (even though it says to use it)

20. KingGeorge

I think it only tells you that mathematica doesn't have the programming.

21. SithsAndGiggles

Hmm... power series? With WA, adding more terms to the cosine series makes the limit alternate between infinity and complex infinity... Not sure what that tells us.

22. tkhunny

Can it converge if it is only intermittently continuous? This is not some countably infinite number of discontinuities (where some Lebesgue measure might provide consistent results). In the logarithm form, there are giant gaps. This is really why introducing the logarithm is a bad idea. It massively modifies the Domain. If you can tell me there is a big number, M, where $$cos(x/2) > 0\;for\;x > M$$, then the logarithm is of no concern. There is no such number M.

23. KingGeorge

This limit can't possibly exist. Since cosine regularly alternates between $$\pm1$$, no matter what value of $$x$$ we're currently looking at, we can get a larger value such that $$\cos(x/2)$$ is 1. Powers have no effect on this, and so you can always get a larger value such that the function outputs 1. But you can also get a larger value whose output will give you 0. So there can't possibly be a limit.

24. tkhunny

Sequence: $$(\cos(4n\pi))^{n^2}\;for\; n\in\mathbb{N}$$ It's just ones (1s).

25. DominicNg

I believe the question is not properly phrase. As x tends infinity, cos(x/2) and sin(x/2) is not defined.