Why is this allowed?

still 0/0 retard

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.

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

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

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

git gud

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

Go.

it's not

>cancelling zeroes
L0Lno fgt pls
Lrn2division

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

This is correct.

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

This is a moron.

OP, a video juuuust for you,

youtube.com/watch?v=YTKoob7m3DM

(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.

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.

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

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'

When is 1-1=1?

retard

Just call it an "imaginary" number.

math's a hoax

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

Since when is that ever is the integral of cosine?

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