Prove or disprove the existence of a limit of the function $\displaystyle \lim_{x\to x_0} f(x)$

Question

Prove or disprove the existence of a limit of the function $\displaystyle \lim_{x\to x_0} f(x)$

$f(x)=\begin{cases} 0 , & x\in\Bbb R\setminus \Bbb Q\\ \sin|x| , & x\in \Bbb Q \\ \end{cases}$

For

1. $x_0\in \{\pi n ,\ n\in \Bbb Z \}$
2. $x_0\in \Bbb R \setminus\{\pi n ,\ n\in \Bbb Z \}$

The way I understood it is we need to show continuity, here 1. always go to zero because it's irrational ($\pi n \notin \Bbb Q$) and 2. go to zero and to sin|x| because it could be irrational sometimes.

I tried to use the Cauchy definition of a limit but I don't really get anywhere...

$$\forall \epsilon\gt0:\exists\delta\gt0:\forall x:|x-x_{0}|\lt\delta\rightarrow |f(x)-L|\lt\epsilon$$

Using Heine definition is a dead end too.

Any help would be appreciated.

NOTE: we can't use integration/derivation/L'hopital/Taylor's because we haven't covered those (i.e. "no calculus").

Answer 1

Notice that any neighbourhood of $x_0$ contains both points of $\mathbb Q$ and points of $\mathbb R\setminus \mathbb Q$. Since both functions defining $f$, namely the function $0$ and the function $\sin |x|$ are continuous, they must have the same limit in $x_0$ if you want that $f$ has a limit in $x_0$. Hence you must check whether: $$ \sin |x_0| = 0. $$

Answer 2

Since the first question is a bit harder (in my opinion) i will solve that one.

Theorem: The $ \epsilon - \delta$ definition for continuity of a function is equivalent to the following definition:

$f(x)$ is continous at $x_0 $ if and only if for every sequence $(x_n)$ such that $ \lim_{ n \to \infty} x_n = x_0$ we have $\lim_{ n \to \infty} f(x_n) = f(x_0)$

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Let us take a look at 2 cases:

OCase 1. Let $(t_n)$ be a sequence of rational numbers such that $t_n \to 2k \pi$ for some $k \in \mathbb{N}$ then obviously $\lim_{n \to \infty} \sin |t_n| = 0$, (show why, this is a textbook example).

OCase 2. Let $(t_n)$ be a sequence of irrational numbers such that $t_n \to 2k \pi$ for some $k \in \mathbb{N}$ then obviously $\lim_{n \to \infty} \sin |x_n| = 0$ (because if $t_n$ is irrational we have $f(t_n) =0$ for all $n$ by definition and hence $\lim_{ n \to \infty} f(t_n) =0$

Now for an arbitary sequence $(x_n)$ there are 3 cases (why?):

Case 1. $(x_n)$ has an infinite number of rational and irrational numbers.

In this case, we can "split" $(x_n)$ using subsequences. Let $(r_n) \in \mathbb{N}$ be the sequence such that $x_{r_n} \in \mathbb{Q}$ and $(i_n) \in \mathbb{N}$ a sequence such that $x_{i_n} \in \mathbb{R}/\mathbb{Q}$. Now what happens with $f(x_{r_n})$ and $f(x_{i_n})$ when $n \to \infty$? If all subsequences of a sequence $(x_n)$ have the same limit then what can we say about the limit of the original sequence $(x_n)$?

Case 2. $(x_n)$ has an infinite number of rational numbers (but finite number of irrational).

Hint: Here there must exist $n_1$ such that for all $n>n_1$ we know that $x_n$ is rational. Then just use OCase 1 to come to a conclusion.

Case 3. $(x_n)$ has an infinite number of irrational numbers (but finite number of rational numbers).

Can we approach this part similarly to Case 2?

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The second part of your problem should be easier. Just take a look at Case 1, assume that the limit exists and come to a contradiction.

As a much more interesting problem which uses the same idea, check this one out:

http://en.wikipedia.org/wiki/Thomae%27s_function and it's informal proof.