Thread: 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??!?!


thanks! please do asap

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 and exist then 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 (that is, assume it exists).


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 .