You still haven't learned your prior lesson, have you? Is \(\frac{sinx}{x}\) when \(x->0\) undefined?

As is, they are undefined. If it is a result of a limiting process (like sinx/x as x -> 0), then there are procedures (L'Hopital's rule) to handle it.

Infinity is not a member of the Reals and therefore it is meaningless to try to do algebra with it unless you define the relevant rules which extend the Reals appropriately. Similarly 0/0 is meaningless because it represents 0 * (1/0), ie when you divide by X you actually multiple by the number which when multiplied by X gives 1. There is no Y such that Y*0 = 1 in the Reals and therefore 1/0 is no more a valid expression than elephant/cheese = table.

Careful, Saint is easily confused. What eram said true: zero/zero is undefined. \(\lim_{x\to 0}\frac{\sin x}{x} = 1\) \(\frac{\sin 0}{0}\) is undefined. \(\lim_{x\to 0}\frac{\sin x}{x}\) is not the same as \(\frac{\sin 0}{0}\).

Exactly, which is why if you define a function \(f(x) = \frac{\sin x}{x}\) you have to define the value of f(0) separately. If you decide f(0) = 1 then that fits with the limit but that isn't required.