Homework Help: Evaluating limit question

1. Oct 15, 2011

appplejack

1. The problem statement, all variables and given/known data
Evaluate the following limit
lim x->0 f(x) = 2, x rational, -2, x irrational

2. Relevant equations

3. The attempt at a solution
I tried to draw the graph of this question but couldn't. Do you think I have a function that oscillates between 2 and -2 violently?
My best guess is that the limit doesn't exist but I want to have better explanation than mine.

Last edited: Oct 15, 2011
2. Oct 15, 2011

Matterwave

That looks like a slightly modified Dirichlet function to me. Between any 2 rational numbers is there an irrational number? And between any irrational number is there a rational number?

Also, what is this limit to? Limit as x goes to what?

3. Oct 15, 2011

appplejack

my apology. x->0.
There's no irrational number between 2 rational numbers. Vice versa no?

4. Oct 15, 2011

Matterwave

er...

Does your last statement make sense?

Between 1 and 2 is there an irrational number?

5. Oct 16, 2011

Mr.Miyagi

When you suspect that a limit does not exist, you can make use of the sequential criterion for limits. The criterion loosely states a limit $\lim_{x→c}f(x)=L$ exists if and only if for every sequence $(x_n)$ that converges to $c$ it follows that $\lim_{n→∞}f(x_n)=L$.

So if you want to prove that the limit does not exist you can do so by showing that there exist two sequences $(x_n)$ and $(y_n)$ that converge to $c$ and satisfy $\lim_{n→∞}f(x_n)\neq \lim_{n→∞}f(y_n)$.

Does this make sense to you? Can you construct two sequences that would work in your case?

6. Oct 17, 2011

appplejack

Sorry. Could you explain it easier?

7. Oct 17, 2011

gb7nash

For a limit to exist, every time we make a sequence approach 0, the corresponding function values must approach the same number.

For example:

If I choose a sequence of rational numbers: 1,1/2,1/4,...

We get: f(1),f(1/2),f(1/4),...

So we get 2,2,2,... and this approaches 2.

However, if we look at e,e/2,e/4,...

We get: f(e),f(e/2),f(e/4),...

So we get -2,-2,-2,... and we get -2.

Since these are two different values, the limit does not exist.