# 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.