If and exist then exists. That is a theorem.

No2. In the formal definition of limits: lim x->c f(x) = L, if for every E>0 there is a d>0 such that IF 0<|x-c|<d THEN |f(x)-L|<E

can you switch the inequalities after the IF and THEN, and does the definition still hold?

Assume that (that is, assume it exists).3. TRUE or FALSE

f(x) = 1 if x is rational

= 2 if x is irrational

the limit as X-->0 exists for f(x)

im not sure about this one cause of whether the right and left hand limits meet and if so, at 1 or 2??!?!

If then pick so that .

This means there is so that if .

Therefore for .

But this is a contradiction since .

If then pick so that .

Repeating a similar argument will show and this is a contradiction too.

Therefore if it exists. Using a similar argument as above we can show . This forces . But then this means there is so that and .

Thus, there is so that .

Thus, .

This is a contradiction beceause does not take values strictly between and .