If I have an equation such as y = (3x(1-x))/(1-x) this simplifies to y = 3x?

but if x = 1 the first equation is undefined though for the second y = 3

could somebody please explain what is going on here?

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- December 29th 2012, 06:15 PMkinhew93Confused about undefined values
If I have an equation such as y = (3x(1-x))/(1-x) this simplifies to y = 3x?

but if x = 1 the first equation is undefined though for the second y = 3

could somebody please explain what is going on here? - December 29th 2012, 06:21 PMskeeterRe: Confused about undefined values
the function has a point discontinuity (a "hole") at the coordinates (1,3)

- December 29th 2012, 11:27 PMDevenoRe: Confused about undefined values
the simplest example is:

y = x/x

for every x EXCEPT 0, we have y = 1. at x = 0, y is undefined.

it seems like we ought to just be able to "get around this" by simply using THIS function instead:

y = x/x, for x ≠ 0

y = 1, if x = 0

which is what we really want.

a while ago (around 450 years, give or take) some clever guys discovered a way to deal with this "pesky vanishing 0 in the denominator"...use LIMITS!

the idea is to use what x/x is "heading to" at x = 0, without letting x actually "touch 0". since this is clearly 1 (in this example), we can "plug the hole" obtaining the "defined everywhere" function we really want, and not some bad one with a point of "un-definition".

unfortunately, limits turn out to be, in general, kind of complicated (this particular one isn't so bad). there's these things called epsilons and deltas which can make your head hurt (mathematicians love fancy names for "itty-bitty numbers").

what you have discovered (albeit unwittingly) is that:

even though 0/0 makes no sense at all.