Crazy limit

>>8292306
I know wolfram alpha use to have problems computing the limit of x^x, this sounds like a computation problem.
Try plotting the function to test it

>>8292316
Google won't graph it and I don't have a graphing tool.

I guess it is a mistake though, it does not make sense.

It does make sense
x^x approaches 0 as x -> - infinity, as the exponent approaches 1/infinity (due to the exponent being negative) therefore x^x -> 0 as x-> -infinity
The rest is trivial
sin(0)=0
x^0 = 1

It's one OP

>>8292324
My nigga.

>>8292324
^^What he said, how is this confusing OP?

>>8292324
Ok, I think I did it this time. I broke the wolfram.

Take a look at

http://www.wolframalpha.com/input/?i=limit+as+x+approaches+-infinity+of++x%5Esin(x%5Ex%5Ex)

Given that the limit of x^x^x is also one then that does not make sense!

Why is that limit infinity?

Even wolfram says that the limit of x^x^x is 0

http://www.wolframalpha.com/input/?i=Limit+as+x+approaches+-infinity+of+x%5Ex%5Ex

>>8292346
I can explain the first one:
x<0 therefore x^x^x = x/(x/x) = x
x^sin(x) must always be between 1/x and x (as sin(x) is always between -1 and 1)' therefore lim x-> infinity is +infinity

The discrepancy arises due to how wolfram alpha chooses to handle infinite values, a sine of an infinite value could be any thing between 1 and -1, (it chose to make it equal to one in your first case).
However, in your second case, the expression was not within any bounded function, so I think wolfram just decide to make the second one equal to zero.

Any better explanations are welcome

>>8292361
>x<0 therefore x^x^x = x/(x/x) = x
what the fuck is this thread

>>8292324
anon that's not a proof and you know it
>the rest is trivial
you've revealed yourself, first year

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>>8292324
[math]-\infty^{0} = 1[/math]
you funny guy I kill you last

>>8292324
I dont think its that simple. The function x^x as a real valued function isnt fully defined for any interval (-inf, -N)
If you look up the graph for x^x they dont show it past the left side of the y-axis. There is a reason for this.

How is nobody pointing out that

(-s)^r

with s>0, for any non-integer r, is not quite well behaved.

The limit of x^x for some negative number will naturally give something odd, if not undefined.

If you consider
abs(x)^sin(abs(x)^x)
instead of
x^sin(x^x)

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>>8292385
Refresh the thread and dont steal my thunder, kid

>>8292380
why not? N^0 = 1 because the zero means you aren't multiplying anything by anything,leaving the multiplicative identity.

A bit like how 0!=1

>>8292394
God why is everybody so retarded itt
except >>8292385

>>8292346
You can't take parts of a limit separately unless the limit exists for both parts individually.
So it's just a coincidence that lim(x^sin(x^x)) was the same as lim(x^lim(sin(x^x))) as you've been told to evaluate it in this thread. (I write just lim to mean limit as x approaches -infinity for convenience.)

One correct way to evaluate your first limit is
lim(x^sin(x^x)) = e^lim(x sin(x^x)) = e^(lim(x^(x+1)) - lim(x^(3 x+1)/6) + ...) = e^(0 - 0 + ...) = 1
where I haven't shown that lim(x^(a x+1)) = 0, but it's not particularly difficult to.

It's more complicated for your other two examples but the gist is that
lim(x^(x^x)) does exist
but
lim(x^((x^x)+1)) doesn't which shows up when you expand the sin.

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>>8292306

>>8292405
So obviously they are implicitly taking |x|^x, because the expression is nonsensical as a real valued function for most negative numbers

>>8292401
>lim(x^sin(x^x)) = e^lim(x sin(x^x))
seriously what?

>>8292401
Whoops it should be log(x) not x in the exponential but it's a similar same idea probably.

>>8292418
>it's a similar same idea probably
yeah brah just take [math]log(-\infty)[/math]

>>8292400
I was merely portraying a façade, you see, my post was not made in earnest! You have become ensnared in yet another one of my devilish schemes, for you see... I am not merely a dim witted fool you have taken me to be, rather, that is merely a mask I have worn in order to lure you into my escarpment. This disguise allows me to strike again, and again, with no repercussions. I will be seeing you again very soon, my dear friend. As you shall fall prey to yet another of my devilish schemes. For I am the witty banterster!

>>8292421
It's log(x) x^x not just log(x) on it's own.

>>8292423
Thunderation! It appears I have been coaxed into a snafu once again.

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>>8292427
yeah brah just take the limited expansion of the log function at -infinity

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Did I type it in correctly?

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>>8292435
I didn't see the negative there.

>>8292438
>that pic
ok I'm fully triggered now
good job baiters

>>8292449
Please see
>>8292423

>>8292434
log(x) ~ log(x+1) - log(1+1/x) as x goes to -infinity
so
x^x log(x) ~ (x^(x+1) - (x^(x+2)/2 + ...) - (x^(x-1) - x^(x-2)/2 + ...)
and x^(x+a) goes to 0 as x goes to -infinity as already discussed

This function is complex valued in this domain. Thats problem number 1. And also probably problem 2, 3, and 4.

>>8292475
>Thats problem number 1
Problem number 1 is it's not fucking defined moran.

>>8292478
No its defined by analytic continuations of all functions there, just not defined on the reals. And Moran is my father, call me Iodit.

>>8292385
>>8292383
Not exactly pertinent to the limit, but is it completely undefined, or does it depend on the space you're working in?

>>8292486
The function x^x, which is the stickler here, is defined properly by the exponential for all x>0. At 0, it is completely undefined, and for most negative x it requires an analytic continuation. It is defined by the elementary definition if x is a negative integer, if x is of the form of a negative even integer over a odd integer, or if x is any rational that is a quotient of odd integers. However it is undefined (on reals) for anything else, like if x=(-1/2). Thats the gist of it maybe I forgot one more case.

>>8292324
>x^x approaches 0 as x -> - infinity

how did no one point this out

>>8292555
actually people have pointed out that was wrong at least 4 times if you read the thread

>>8292555
My mistake.
It should have been as abs(x^x) -> -infinity.

>[math]\infty[/math]
I expected /sci/ to know better

>>8292571
Quick question:
Does [math] \lim_ {x \to \infty} (-2)^(-x) \neq 0 [/math] ?

>>8292321
https://www.desmos.com/

>>8292641
>(-2)(-x), i.e. 2x
take your homework to >>>/hm/

>>8292438
SEE

I AM NOT CRAZY. THIS SHIT IS WEIRD.

GOD IS NOT IN HIS HEAVEN. MATHEMATICS IS NOW A FREE FOR ALL, A TRUE ANARCHY. IF WOLFRAM CAN'T PROVE IT THEN NO ONE CAN.

Pretty much any function where sine contributes greatly will converge at 1.

>>8292641
-inf

>>8292790
Sin(x)?

>>8292795
Sry sin(x) diverges

>>8292383
No one said that x was a real. In the context of this limit, the only set that makes sense is the integers. The limit of x^x when x goes to minus infinity as am integer is indeed zero. More precisely, it is an alternated sequence that converges to zero.

-(inf)^(-(inf)) = -1/( (inf)^(inf) ) = 0
sin(0) = 0
-(inf)^0 = 1
Why? Becase, 0 means n-n. Okey.
-(inf)^(n-n) = inf / inf = 1

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>>8292660

what the fuck desu

It's divergent, retards

>>8292509
isn't [math]\frac{1}{2}^\frac{1}{2}=-\sqrt{2}i

>>8292361
Well, I have access to some of the intermediate steps, so let me explain.

I began by seeing what the steps are to [math] \lim_{x\to-\infty} \sin(x^x) [/math]. They are:
[math] \sin x [/math] is a continuous function, so therefore:
[eqn] \lim_{x\to-\infty} \sin(x^x) = \sin\left(\lim_{x\to-\infty} x^x\right) [/eqn]
[eqn] = \sin\left(\exp\left(\lim_{x\to-\infty} x \ln x \right)\right) [/eqn]
[eqn] = \sin\left(\exp\left( \lim_{x\to-\infty}\left( x \right) \cdot \lim_{x\to-\infty}\left( \ln x \right)\right)\right) [/eqn]
[eqn] \lim_{x\to-\infty} \ln x = \infty [/eqn]
[eqn] \therefore \lim_{x\to-\infty} \sin(x^x) = \sin e^{-\infty \cdot \infty} = 0 [/eqn]

>>8292306
nigga your function isn't even well defined
|x|^sin(|x|^x) does approach 1 obviously but x^sin(x^x) doesn't

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Back at my home computer, and with Mathematica at my side, let me try to sketch this out.

First:
[eqn] x^{\sin x^x} = \exp\left(\sin\left(e^{x \ln x}\right) \ln x \right)[/eqn]
Then we go:
[eqn] \lim_{x\to-\infty} \left( \exp\left(\sin\left(e^{x \ln x}\right) \ln x \right) \right) = \\
\exp\left( \lim_{x\to-\infty} \left( \sin\left(e^{x \ln x}\right) \ln x \right) \right) =\\
\exp\left( \lim_{x\to-\infty} \left( \sin\left(e^{x \ln x}\right) \right)\cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left(\lim_{x\to-\infty} \left(e^{x \ln x}\right) \right)\cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left( \exp\left( \lim_{x\to-\infty} x \ln x\right) \right) \cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left( \exp\left( \lim_{x\to-\infty} \left( x \right) \lim_{x\to-\infty} \left( \ln x \right) \right) \right) \cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left( \exp\left( \lim_{x\to-\infty} \left( x \right)\right)^{\lim_{x\to-\infty} \left( \ln x \right)} \right) \right)^{\lim_{x\to-\infty} \left( \ln x \right)}
[/eqn]
Now we only have to calculate two actual limits:
[eqn]
\lim_{x\to-\infty} x = -\infty \\
\lim_{x\to-\infty} \ln x = \lim_{x\to-\infty} i\pi + \ln (-x) = \infty
[/eqn]
Plugging those back in:
[eqn]
\exp\left( \sin\left( \exp\left( -\infty \right)^{\infty} \right) \right)^{\infty}
[/eqn]
And now collapse it inward:
[eqn]
e^{-\infty} = 0 \\
0^\infty = 0 \\
\sin 0 = 0 \\
e^0 = 1
[/eqn]
The problem is that the last step, [math] 1^\infty [/math] is an indeterminate expression. Mathematica will evaluate [math] \lim_{x\to\infty} 1^x [/math] as 1, but that's actually subtly wrong. See http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form for an explanation.

>>8297654
> [math]\lim\limits_{x \to -\infty} \ln x = \lim\limits_{x \to -\infty} i\pi +\ln(-x) = \infty [/math]
breaking news, tripfag is retarded

>>8297668
It's called the principal complex logarithm. It works here because we have already constrained that [math] x < 0 [/math]. The choice of [math] i\pi [/math] is the branch cut.

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>>8292324

>>8292324

1/(-∞^∞) = 0?

w8 m8, is the generally agreement in conventional academia that infinity to the infinity is infinity?

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>>8292380
>slow dead

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>>8292306
[math] n \to k \quad k = -1 \\ "-1^-1"[/math]

please kill yourself, this makes me hurt