/sci/ - Science & Math

[Anonymous No.11674720]

you should be able to solve this

[Anonymous No.11674721]

>>11674720
50/50 it either happens or it doesn’t.

[Anonymous No.11674724]

>>11674720
-1/12

[Anonymous No.11674726]

>>11674720
The answer is that "less than 1" isn't the same as "1".

[Anonymous No.11674728]

>>11674720
1

[Anonymous No.11674746]

1

[Anonymous No.11674749]

>>11674746
>proof by graphical inspection
Heinously based

[Anonymous No.11674750]

1/infinity is 0 and infinity ^ 0 is 1 so 1 so this is some basic high schooler math

[Anonymous No.11675665]

>>11674728
>>11674746
>>11674750
Wrong.
Let us consider the function [math]x^(1/y)[/math]
Then we can try to find the limit as x and y both go to [math] \infty [/math]
[math] \lim\limits_{(x,y)\to(\infty,\infty)} x^{1/y} [/math]
We first consider the following limts:
[math] \lim\limits_{x\to\infty} \lim\limits_{y\to\infty} x^{1/y} = \lim\limits_{x\to\infty} x^0 = 1 [/math]
[math] \lim\limits_{y\to\infty} \lim\limits_{x\to\infty} x^{1/y} = \infty [/math] (since [math] \frac{1}{y} [/math] is positive)
Therefore the limit does not exist and the value of such an expression depends on how you approach it.

[Anonymous No.11675674]

>>11675665
graph dont lie

[Anonymous No.11675678]

>>11674720
0.99999999999999999999999999999

[Anonymous No.11675699]

>>11674720
inf^(1/inf) = inf^(0) = 1

[Anonymous No.11675710]

>>11674720
put it in limit form what you mean by infinity

[Anonymous No.11675713]

>>11675674
Graph is for [math] x^{1/x}[/math] not for [math] x^{1/y} [/math] (two varibles). If you don't belive me go graph it youreslf in 3D space and see that it has the exact behavior I described retard.

[Anonymous No.11675736]

>>11675713
graph dont lie

[Anonymous No.11675753]

>>11675736
Go graph [math] x^{\frac{1}{\lnx}} [/math] ([math] \lnx\to\infty [/math] when [math] x\to\infty[/math] so it is a [math] \infty^{\frac{1}{\infty}} [/math]) and see that it is equal to e, not 1 retard.

[Anonymous No.11675754]

>>11674720
Im a retarded 19 yo undergraduate w/ 0 papers to my name and lots of opinions

[Anonymous No.11675764]

>>11675753
>not 1 retard
Correct, you're retarded enough for two people.

[Anonymous No.11675763]

>>11674720
Lim(x^(-1/x)) = lim exp(ln(x^(-1/x))) = lim exp(-1/x)^ln(x) = 1

[Anonymous No.11675799]

>>11675665
Retard

[Anonymous No.11675814]

>>11675665
What is 1 divided by infinity? 0. What is infinity to the 0 power? 1 easy pz. I’m not sure what sort of schizo math this is.

[Anonymous No.11675828]

>>11675764
>>11675799
He's right is what he is. What is it about the OP that makes you interpret this as the limit of x^(1/x)? There is nothing in the OP to indicate that?

[Anonymous No.11675840]

>>11675828
Why do you interpret as x ^ (1/x)
It’s infinity ^ (1/infinity)
We can both accept 1/infinity is 0
Infinity ^ 0 is 1 I’m not sure what sort of schizo math they are trying to pulll.

[Anonymous No.11675908]

>>11675753
Don't be stupid... Evaluate [math]\frac{x}{x}[/math] and [math]\frac{x}{x^2}[/math] as [math]x \rightarrow \infty[/math]. If everything goes to infinite it is the same thing, right?

[Anonymous No.11675955]

>>11675840
Well x^(1/lnx) is also infinity^(1/infinity) and it's limit is e

>>11675908
The first is 1, the second is 0. They both evaluate [math]\frac{\infty}{\infty}[/math] which depends on how you approach it too.

[Anonymous No.11675976]

>>11674724
Pop math retard a show your proof

[Anonymous No.11675989]

>>11674720
undefined
https://www.wolframalpha.com/input/?i=inf%5E%281%2Finf%29

[Anonymous No.11675991]

graph don't lie

[Anonymous No.11676009]

>>11675989
wolfram lies

[Anonymous No.11676019]

>>11675955
imagine getting gtfo by >>11675908
but still squirming around like an oily pig

[Anonymous No.11676037]

>>11675955
it doesnt matter how you approach it if you graph x/x^2 it will go to 0 not some other crazy schizo shit you are trying to say it will go to if you start saying "well infinity equivalent to infinity"

[Anonymous No.11676099]

>>11676019
>>11676037
Reatrds it matters how you approach [math]\frac{\infty}{\infty}[/math]
If you approach it with x/x you get a different result from when you approach it with x/x^2. That was my original point. The same is true for inf^(1/inf). Imagine being this reatrded.

[Anonymous No.11676104]

>>11676009
ok schizo

[Anonymous No.11676161]

>>11674720
Lets look at [math]\lim_{x\to\infty} \text{ln}(x^{1/x}) [/math] instead.
We then have [math]\text{ln}(x^{1/x})=\frac{\text{ln}(x)}{x} [/math] which by by L'Hopital approaches
[math]\lim_{x\to\infty}-\frac{1}{x^3} [/math] as its limit. So taking the exponential we see [math]\lim_{x\to\infty} x^{1/x}=1 [/math]

[Anonymous No.11676198]

>>11676161
[math] \displaystyle
\lim_{x \to \infty} 1^x = 1 \\
1^ {\infty} ~~ undefined
[/math]

[Anonymous No.11676218]

>>11676198
How can that be undefined, if 1 is 1 literally in every power. I would rather agree that 1^0 is undefined, but I saw the graph.

[Anonymous No.11676224]

>>11676218
https://www.wolframalpha.com/input/?i=1%5Einf

[Anonymous No.11676288]

>>11676218
You are thinking about the function 1^x which is 1 for all x and (therefore) has limit 1 as x goes to [math]\infty[/math]. But [math]1^\infty[/math] is not the limit of that function. It is the limit of any function [math](f(x))^x[/math] where the limit of f is 1 as x goes to [math]\infty[/math]. It depends on how your function approaches 1. For exapmle (x+1)/x approaches 1 but ((x+1)/x)^x doesn't go to 1 (even though it's [math]\frac{1}{\infty}[/math]). This is true for other values like [math]\frac{\infty}{\infty}[/math] and [math]0^0[/math] but not for all such expressions (i.e. [math]\frac{1}{\infty}[/math] is always 0). I've been trying to explain this in this thread but apparently nobody (with few exceptions) understands this stuff here. (And they are calc 1 stuff)

[Anonymous No.11676293]

If you don't understand basic fucking limit issues you need to leave this board.

[Anonymous No.11676295]

>>11676288
>(even though it's [math]\frac{1}{\infty}[/math])
Meant (even though it's [math]1^{\infty}[/math])

[Anonymous No.11676354]

>>11676218
but where is that operation used?

[Anonymous No.11676406]

>>11674720

In what case would I actually need to solve this though
What does an infinite root even represent

[Anonymous No.11676456]

>>11676218
>I would rather agree that 1^0 is undefined
Then you are retarded. You don't even need limits to show that.
Consider [math]a \neq 0[/math] and [math] n > 0[/math]
[math]a^0=a^{n-n}=\frac{a^n}{a^n}=1[/math]
Therefore [math]a^0=1, \, \forall a \neq 0[/math]

[Anonymous No.11676467]

>>11674720
1

[Anonymous No.11676502]

>>11674720
Nooo, you just can't get the non number root of a non numberinooo

[Anonymous No.11676553]

>>11676456
I don't deny that it's 1, I only say that 1^inf = 1 is even more obvious.