Why is this allowed?

>>8650080
[math](1^{2}+1)(1-1) = 1^{2}*1 -1^{2}*1+1*1-1*1 = 1-1+1-1 = 0[/math]

that's not allowed

>>8650124
/thread

>>8650080
incorrectly cancelling zeroes, mathematical and logical error.

omg fuck you

god damit
FOR FUCKS SAKE

>>8650080
What's supposed to be the cleverly bad interpretation? How could someone incorrectly think it is 1/2?

[math]\frac{1-1}{1^{2}-1}[/math]

>>8650080
this is not allowed you stupid fuck

File: 1455656640364.png (38KB, 365x214px) Image search: [Google]

38KB, 365x214px

>>8650080
still 0/0 retard

>>8650124
>>8650190
>>8650221
>>8650421
>>8650441

let the firts one be an x

(x-1)/(x^2 - 1), then your pic is equal to f(1), but it doresnt exist, so take the limit of (x-1)/(x^2 - 1) with x going to 1 and you get x/(x+1), set x = 1 amnd you get 1/2, QED.

>>8650690
yes, you brainlet, the limit, not the actual value

>>8650714
oh honey, come back and replay again after you've taken an analyses course.

>>8650782
Anal is for fags, come back when you aren't a fag

File: bern.gif (3KB, 583x101px) Image search: [Google]

3KB, 583x101px

>>8650080
git gud

>>8650080
Because 0/0 can equal whatever the fuck you want it to be.

>>8650865
Go.

>>8650080
it's not

>>8650190
>cancelling zeroes
L0Lno fgt pls
Lrn2division

>>8650714
>yes, you brainlet, the limit, not the actual value

This is correct.

>>8650782
>oh honey, come back and replay again after you've taken an analyses course.

This is a moron.

>>8650080

OP, a video juuuust for you,

https://www.youtube.com/watch?v=YTKoob7m3DM

>>8650690
(1) That function is not equvilant. 1^2 = 1, therefore it would be (x-1)/(x-1).

(2) 1^2 = 1, therefore you could argue that (1^2-1) = (1+1)(1-1) = (1+1)(1^2 - 1) = (1+1)(1+1)(1^2 - 1). You can then "argue" it equals whatever you want with your reasoning:
[math]
\frac{1 - 1}{1^2 - 1} = \frac{1 - 1}{1^{4} - 1}
[/math]

Now applying your indescribably bad math:

[math]
= \frac{(x - 1)}{(x^{4} - 1)}\\
= \frac{(x - 1)}{(x^2 - 1)(x^2+1)}\\
= \frac{(x - 1)}{(x + 1)(x-1)(x^2+1)}\\
= \frac{1}{(x + 1)(x^2+1)}\\
\lim_{x\to1} \left(\frac{1}{(x + 1)(x^2+1)}\right) = \frac{1}{4}
[/math]

You complete retard.

>>8650949
you are right that 1^2 = 1^4 in value, but 1^2 != 1^4 AS A DEFINITION, so you cant just change them. Heres an example anyone with calc 1 would have heard before

let f(x) = (x-1)/(x-1) and g(x) = 1, from this we know that f(x) is not equal to g(x), because one is not defined at 1 and one is, but the VALUE of the 2 are the same when you apply proper limits. same with 1^2 and 1^4, the VALUE is the same, but the definition is not.

>>8650080
If you're taking the limit it's correct, but if you're dividing numbers then it's not correct.

difference of two squares doesn't work of a==b

>>8650080
hmm thing here is that not only denominator but also numerator are zero, with that (bait) approach you can also 'show/prove' that division by 0 can be -0.5 , 0.5 ,2 , -2 , 1 , -1 ,0, infinity , ... etc. when dividing by zero depending form the direction you approach the zero, your result can be pretty much anything you want, hence we 'dont divide by zero'

>>8650080
When is 1-1=1?

>>8652369
retard

Just call it an "imaginary" number.

>>8652497
math's a hoax

>>8652378
>take 1 away from 1
>now there's nothing there
>I'm the retard for not pretending there is
Nah, go fuck yourself

>>8650789
Since when is that ever is the integral of cosine?

>>8650080
I think there's a missed negative and 1 in the answer.
Shouldn't it be -1/12
>KEK

You're supposed to simplify as the first step. Even if you expanded it, you'd need to work out the parentheses before dividing (1-1) over (1-1). Which events in 0/0