Math question.

[joonkimdds]

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i was just going over every math materials i have and here r couple of questions i don't get.

what's 0 over 0?

in the function, f(x) = x/[x(x-1)] , x=0 and x= +1 makes denominator =0.
but there is only 1 vertical asymtote when x=+1. why isn't it when x= 0 ?
if we put 0 for a value of x, it's 0/0 and that means there is Y when X=0.
The meaning of vertical asymtote is that there is no value at that point.
so I think x=0 also makes vertical asymtote....where did i go wrong?
one of my friends told me that when i cancel out up and bottom like x=0, they r called removable discontinuity, not vertical asymtote, but I don't understand the difference.


I feel like i am asking a calculus problem but i believe i learned it in pre calculus so i just thought that it's good to go over little bit of basic calculus.

 

[trowell460]

there is no y when x=0 y is undefined

 

[joonkimdds]

there is no y when x=0 y is undefined

so when y is undefined...shouldn't it be vertical asymtote?
vertical asymtote means there is no value when x is some value.

 

[coolmoe_33]

if f(x) = x/[x(x-1)] then the x in the numerator cancels with the x in the denominator and u end up with only 1/(x-1), which means only one vertical asymptote at x= 1.

 

[UW09]

As far as I can remember I believe that zero over zero can only be a HOLE...so it's never any asymp.

 

[Pelotari]

i was just going over every math materials i have and here r couple of questions i don't get.

what's 0 over 0?

in the function, f(x) = x/[x(x-1)] , x=0 and x= +1 makes denominator =0.
but there is only 1 vertical asymtote when x=+1. why isn't it when x= 0 ?
if we put 0 for a value of x, it's 0/0 and that means there is Y when X=0.
The meaning of vertical asymtote is that there is no value at that point.
so I think x=0 also makes vertical asymtote....where did i go wrong?
one of my friends told me that when i cancel out up and bottom like x=0, they r called removable discontinuity, not vertical asymtote, but I don't understand the difference.


I feel like i am asking a calculus problem but i believe i learned it in pre calculus so i just thought that it's good to go over little bit of basic calculus.

If you have a fraction (say 1/2) and multiply numerator and denominator by a number (call it x), you are not changing anything, right? Same thing with f(x) = x/x(x-1).

True, you can argue that x cannot be zero, but I can also argue that x TENDS to zero (remember limits?). Since x will be close to zero (and not equal), I can cancel the X's and my function really is f(x) = 1/(x-1). This looks exactly like the function f(x) = 1/x, just shifted to the right by one. Your vertical asymptote is at x=1. Your horizontal asymptote is the "Y" axis (i.e. at y=0).

Hope this helps.

 

[busdriver]

I always thought that anything over 0 equals infinite (the vertical asymptote when x=1) EXCEPT when it's 0/0. then it just equals 1. so that's probably why there's no asymptote for x=0. whether i'm right or wrong, i don't think you'd see something like this on the DAT. and if there was, it wouldn't be this strange...