??? Thus far if I have seen this in an expression I have just

>Thus far if I have seen this in an expression I have just evaluated it twice, once as +1 and once as -1 and they have both turned out the same.

KEK

>>7965476

I know, I know.

the point is, it worked. but now i dont know what to do.

>>7965450
PRECALC GO AWAY

>>7965450
the limit doesn't exist

File: image.jpg (96KB, 704x391px) Image search: [Google]

96KB, 704x391px

>>7965450

>>7965450
I bet it is 1

-1/12

>>7965450
Try instead evaluating the limit of i^(2x) as x goes to infinity. You can thank me lalter!

>>7966247

How tf is this divergent, it has a set, singular value that never grows.

>>7965450
i'm curious, what did you use it in?

>>7965450
>inf is a even number because it's still whole if you divide it by two
:^)

>>7966303
Negative exponents?

>>7965450
[math]\lim_{x\to\infty}(-1)^{x}=\{-1,1\}[/math]

>>7966303
if you knew what divergent meant then you'd know why. don't use big boy words unless you know what they mean.

>>7966341

It's still always between 1 and -1

Are you on crack

>>7966367
Honestly I don't even know what I was thinking

>>7965483
And somehow these threads always continue past this point.

What's the limit of cos(x) as x goes to infinity?
If you can answer that, you can answer your question.

ye fuckin idiot

(-1)^x = e^(ln(-1)*x) = e^(i*pi*x) = cos(pi*x)+i*sin(pi*x)

the limit of cos(x) as x tends to infinity is undefined
same for sin(x)
YOU FUCKIN IDIOT YOU

>>7965450
[eqn] \limsup_n (-1)^n = 1[/eqn]
[eqn] \liminf_n (-1)^n =-1 [/eqn]

>>7965450
This does not converge, just like sin(x) to x->inf doesn't converge (exactly the same thing, actually))

>>7966303
It diverges because it doesn't stop oscillating, it makes a complex sin/cos wave

>>7965450
The limit does not exist as it does not converge to a specific point. Its ambiguous.

>>7966575

The function itself has an oscillating nature, true. When I said "it" I was referring to the expression in the OP, that has a discrete value, not a divergent one.

File: images_article_2012_09_29_the-limit-does-not-exist.gif (1MB, 630x354px) Image search: [Google]

1MB, 630x354px

>>7965450

>>7966392
1 max and -1 min so average is (1-1)/2 = 0

0 is the limit of cos(x)

>>7966739
Why is the limit the average? Hint: It's not.

>>7966739

But that's fucking wrong. If you're trolling then well played. If you're serious then look at the graph of cosine. And tell me why you think if it goes on infinitely that you think you can find a value for it.

>>7965450
thats an exponential function idiot, it only has values on the complex plane

(-1)^x = e^(i*pi*x)
This is the unit vector in the complex plane and it's moving in circles at a constant rate dx/dt so it has an average value 0 and rms value 0.707

>>7966881
okay but what does that have to do with the limit

>>7966350
Wrong. Think about what happens when x isn't an integer.

>>7966888
I wasn't being serious, but if you assume (-1)^x to be an ac source and use a voltmeter to measure its value, it shows 0.707 V at steady state (as x -> infinity)

>>7966894
x can only be an integer or else it is undefined

>>7966908
correction: x cannot be (1/n) with n being an even number or else you get complex numbers

>>7966912
so since the limit oscillates between 1 and -1 towards infinity it diverges

>>7966900
>>7966881
Although phasor notation is used with rms values so the rms might as well be 1

>>7966908
(-1)^1/2 = i

Isn't it sideways eight imaginary units?