Question

LIMIT!

using L'Hopital's rules, evaluate
lim x appraoches infinity (cos(x/2))^x^2

Answer 1

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

Answer 2

so what I'm say is

you can write
\[(\cos(x/2))^{x^2}=e^{\ln((\cos(x/2))^{x^2})}=e^{x^2 \ln(\cos(x/2))}=e^{\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}}\]

Answer 3

i think i forgot about it

Answer 4

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

Answer 5

wow, seems it is so complicated.

Answer 6

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

Answer 7

@myininaya

Answer 8

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

Answer 9

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

Answer 10

not 0/dne or dne/inf

Answer 11

Oh boy D: Good question..
hmm

So we have a problem since ln(cosx) is undefined over and over as x->infinity, yah?

Answer 12

yah

Answer 13

Grr I dunno >.< maybe lego man knows

Answer 14

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

Answer 15

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

Answer 16

LHops international house of limits

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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.

Answer 22

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.

Answer 23

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.

Answer 24

Sequence: \((\cos(4n\pi))^{n^2}\;for\; n\in\mathbb{N}\)

It's just ones (1s).

Answer 25

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