Mfw he says 0^0 = 0

I think he means [math]0\cdot\[\lim_{x\to\infty} \] \frac 1 x = 0[/math]

Veeky Forums latex interpreter is a mongoloid made fuckfest

You need 0^0 = 1 to write formulas like:

[eqn] e^x = \sum_{k=0}^\infty \frac{x^k}{k!} [/eqn]

Can someone please explain what the fuck I'm looking at?

It's undefined.

see

I think if [math] f [/math] is any continuous function with

[math] \lim_{x\to p} f(x) = 0 [/math], then

[math] \lim_{x\to p} f(x)^{f(x)} = 1 [/math]

In fact,

[math] y^y =1-(1-y)+(1-y)^2-\frac{1}{2}(1-y)^3+\frac{1}{3}(1-y)^4+\left( -\frac{1}{12} \right) (1-y)^5 + O((1-y)^6) [/math]

Another perspective is

en.wikipedia.org/wiki/Empty_product

...

0/0 = 1/100

x^x is not the only way to approach 0^0
you can just as easily approach it from 0^x which will give you a limit of 0
since the double limit has two different values from two different approaches, it is not defined
undergrad math students should know this from multivar calc

It's not like I said "here, that's a proof that 0^0 = 1". It's a perspective on the issue, as I fucking wrote, and like the link to the Empty product, which is about a convention, also shows.

y=0^x

x=2: y=E*0*0=0
x=1: y=E*0=0
x=0: y=E=1
x=-1: y=E/0=undefined
x=-2: y=E/(0*0)=undefined

Where E stands for multiplicative identity, namely, one .(Called it E to make it differ from 1 visually)

Lets look at 0^(x^2) instead of 0^x to avoid problems with division by zero.
If lim x->0 of 0^(x^2) was 0 then f(x) = 0^(x^2) would be identical to f(x) = 0 and it's power series would be all 0's.
But to get its actual power series requires evaluating ln(0) which is undefined. A contradiction so it can't be 0.
Nevertheless if you carry on manipulating ln(0) as if it was a constant (hey, it worked with imaginary numbers!) you get 0^(x^2) = 1 + ln(0) x^2 + ln(0)^2 x^4/2! + ...
Setting x to 0, you're left with just 1.

Someone who knows math explain why this is wrong.

why do you think it's wrong?