ect.

I will accept solutions like 0 and infinity, if they are fully defined (e.a. 0*infty==?).

Edit: I have just thought something:

We can say 3/0=4/0 in a way-3/0 and 4/0 are both indeterminate (if we assume that 2 indeterminates are equal).

I find a case. do you think there is too much difference between enjoying Bill Gates' fortune and owning Warrant Buffet's investment?

Besides, it's not me who proposed infinity is an answer.

Anyway, to Raulito, if you accept some error and do not persuit absolute accuracy, there are accepted solutions.

Otherwise, there is none.

For the limes thingy:

,

If

that doesn't imply:and...again...NO SOLUTION!!!

(If there was some limmy solution, then all math theory would have been contradictory-and this would have been TERRIBLE!!)

No perfect solution.

But accepted solution. If you find a 0 on a calculator display credible for 0. Then there are many answers.

]]>`\mbox{if x is irrational}`

Produces:

]]>Let see what's the limit as x goes to 2.

Here we have 2 different limits: left-limit and right-limit:

See? You can't conclude what is f(x) using only limits.

(and there are more interesting examples: the function

)

::Edited. Changed some latex text.]]>

I think...]]>

If ,

that doesn't imply:

and...again...NO SOLUTION!!!

(If there was some limmy solution, then all math theory would have been contradictory-and this would have been TERRIBLE!!)

Sure.

yeh i realised that about 20 minutes after i posted it, but the thread seemed to have came to an end, so i didnt change it

]]>as x approaches 0, the value of 4/x and 3/x both approach infinity, and become infintisamely different

Anyone care to dispute this limit luca-deltodesco suggests is zero? I do.

]]>