Crazy limit

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

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)

Refresh the thread and dont steal my thunder, kid

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

God why is everybody so retarded itt
except

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.

...

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

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

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

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

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!

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

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

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

Did I type it in correctly?

I didn't see the negative there.

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

Please see

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.

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

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

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

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.

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

how did no one point this out

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

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

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

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

desmos.com/

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

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.

-inf

Sin(x)?

Sry sin(x) diverges

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

what the fuck desu

It's divergent, retards

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

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]

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

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 math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form for an explanation.

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

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.

...

1/(-∞^∞) = 0?

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

>slow dead

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

please kill yourself, this makes me hurt