Math Help - need help with three conceptual calculus problems!

1. need help with three conceptual calculus problems!

ok so 1.

true or false (and provide reasoning, not a full length proof) that
lim as x -> a of f(x)*g(x) can not exist
if lim x->a f(x) and lim x->a g(x) both exist
so its kind of a reasoning to show that the product limit law always works if both limits exist?

2. 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?

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??!?!

2. Originally Posted by stones44
true or false (and provide reasoning, not a full length proof) that
lim as x -> a of f(x)*g(x) can not exist
if lim x->a f(x) and lim x->a g(x) both exist
so its kind of a reasoning to show that the product limit law always works if both limits exist?
If $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist then $\lim_{x\to a}f(x)g(x)$ exists. That is a theorem.

2. 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?
No

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??!?!
Assume that $\lim_{x\to 0}f(x) = L$ (that is, assume it exists).

If $L < 1$ then pick $\epsilon > 0$ so that $L + \epsilon < 1$.
This means there is $\delta > 0$ so that if $0 < |x| < \delta \implies |f(x) - L| < \epsilon$.
Therefore $L - \epsilon < f(x) < L + \epsilon < 1$ for $0 < |x| < \delta$.
But this is a contradiction since $f(x) \not < 1$.

If $L > 2$ then pick $\epsilon > 0$ so that $L - \epsilon > 2$.
Repeating a similar argument will show $f(x) > 2$ and this is a contradiction too.

Therefore $1\leq L \leq 2$ if it exists. Using a similar argument as above we can show $L\not= 1, L\not = 2$. This forces $1 < L < 2$. But then this means there is $\epsilon > 0$ so that $1< L - \epsilon$ and $L+\epsilon < 2$.
Thus, there is $\delta > 0$ so that $0 < |x| < \delta \implies |f(x) - L| < \epsilon$.
Thus, $1 < L - \epsilon < f(x) < L + \epsilon < 2$.
This is a contradiction beceause $f$ does not take values strictly between $1$ and $2$.