Question

This question is buzzing me...

\(\huge \frac{x-x}{x-x}\)

Answer 1

Should the answer be 1 or undefined?

Answer 2

Since have the same numerator and denominator, these two cancels out, leaving 1 as answer...

but \(x-x=0\) and \(\huge \frac{0}{0}\) is underfined.

Answer 3

And the first argument perfectly makes sense! D:

Answer 4

split terms

\[\large \frac{x}{x-x} - \frac{x}{x-x}\]
PEMDASsodivide first

\[\large \infty - \infty\]
so it's indeterminate

Answer 5

And if we work backward...

\(\huge \frac{x-x}{x-x}=1\)

\(1*(x-x)=(x-x)\)

Answer 6

Also:
\[x= a \implies \frac{x-a}{x-a} \ne 1\]

Answer 7

http://www.wolframalpha.com/input/?i=(x-x)%2F(x-x)

Answer 8

Because its still zero over zero from my understanding.

Answer 9

What is the point of replacing \(x\) by \(a\), if \(x=a\)? :|

Answer 10

but according to l'hospital it's 1

Answer 11

Both answers are accepted? o.O

Answer 12

Not its not...
You get:
\[\frac{\frac{d}{dx} (x- x)}{\frac{d}{dx}(x-x)}=\frac{1 - 1}{1 - 1} \ne 1\]

Answer 13

No its not**

Answer 14

Right @eliassaab ?

Answer 15

oh wait lol...i was looking at x-a

Answer 16

I treated \(x-x\) as a term, a term dividing by itself always give 1!

Answer 17

it's a removable discontinuity at 1

Answer 18

zepp...You can't just treat it as a term because its still zero...

Answer 19

it's like saying 2 = 1

Answer 20

:/

Okay then what about this:

Answer 21

literally

Answer 22

\(1*(x-x)=(x-x)\)

This is absolutely right

Then why can't we say that \(\frac{x-x}{x-x}=1?\) :D

Answer 23

It's not correct...(x-x)/(x-x) is NOT EQUAL TO 1

Answer 24

(x-x)/(x-x)=/=1 implies 1(x-x) =/= (x-x)

Answer 25

Even wolfram agrees its indeterminate.

Answer 26

Can you say that (5-3-2)/(5-3-2) is 1?

Answer 27

Wait, so you are saying that

(x-x)/(x-x)=/=1 implies 1(x-x) =/= (x-x)

\((x-x) \ne (x-x)\)?

Answer 28

0/0 is indeterminate. That DOES NOT mean it can't be 1. In this case however, it is not. Look at:
\[\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1\]
That's a 0/0 case that gives you 1.

Answer 29

Oh what a weird paradox xD

Answer 30

I'm saying you're starting from a false assumption to arrive a true conclusion, you can do that if you've ever taken any proof based mathematics or logic...

Answer 31

Umm, I see

Answer 32

Haha, thanks, it's just to see if my proof works (but unfortunately it doesn't). Thanks :)