# How to show that $\lim_{x \to 0} f(x)$ does not exist for a piecewise function involving irrationality?

Let $$f(x) = \begin{cases} 0 & \text{if x is rational} \\ 1 & \text{if x is irrational.} \end{cases}$$

I want to show that $$\lim_{x \to 0} f(x)$$ does not exist. Suppose that $$\lim_{x \to 0} f(x) = L$$ for some real number $$L$$. This means that for all $$\varepsilon > 0$$, there exists a $$\delta > 0$$ such that $$|f(x) - L| < \varepsilon$$ for all $$|x|<\delta$$ with $$x \in \mathbb R$$. Since I wanted to prove the negation, I have to show that for all real numbers $$L$$, there exists an $$\varepsilon > 0$$ such that for all $$\delta > 0$$, there is a number $$x \in \mathbb R$$ such that $$0 < |x|<\delta$$ and $$|f(x) - L|\ge \epsilon$$.

I pick the case that $$L=1$$ for example, and I let $$\varepsilon = 1/2$$ and $$\delta > 0$$. Suppose that $$\delta$$ is irrational. In the case that $$0 < \delta < 1$$, if $$x_0$$ is irrational, then $$f(x_0) = 1$$ and so $$|f(x_0) - L| = 0 < \epsilon$$. This implies that $$x_0$$ cannot be an irrational number that satisfies $$|f(x_0) - L| \ge \varepsilon$$. So $$x_0$$ must be rational. The question is, how can I find a rational number $$x_0$$ such that $$0 < x_0 < \delta$$ for any arbitrary positive irrational number $$\delta$$ strictly less than one?

This has confused me a lot. Is there an easier way to prove that a limit does not exist? I've even tried doing a proof by contradiction, but that brings me to the same result.

• $f(\frac{1}{n})\to 0$ whereas $f(\frac{\sqrt 2}{n})\to 1$ when $n\to \infty$.
– Surb
Commented Jan 29, 2020 at 12:59
• The only thing is that as n approaches infinity, you can't really tell when n is irrational or rational. It's possible that n could be a multiple of $sqrt(2)$, which could make $sqrt(2)/n$ rational.
– Tim
Commented Jan 29, 2020 at 13:02

## 2 Answers

What you wrote doesn't make sense. You wrote “I have to show that for all real numbers $$L$$ […]” and then you stated that $$L=1$$.

Let $$L$$ be an arbitrary real number. Then $$L$$ cannot be equal to both $$0$$ and $$1$$. Suppose that $$L\neq0$$. Take $$\varepsilon=\lvert L\rvert$$. Now, let $$\delta>0$$. The interval $$(-\delta,\delta)$$ contains rational numbers and for each such rational number $$x$$, $$f(x)=0$$. So, we have $$\lvert x\rvert<\delta$$ and $$\bigl\lvert f(x)-L\bigr\rvert\geqslant\varepsilon$$.

The case in which $$L\neq1$$ is similar (take $$\varepsilon=\lvert L-1\rvert$$).

• I was doing a case by case basis, using L as a separate one. My intention was to do the cases when L>1, L=1, 0<L<1, L=0, and L<0. But I'm confused about why you said that L cannot be equal to 1?
– Tim
Commented Jan 29, 2020 at 13:05
• Exactly where did I say that $L$ cannot be equal to $1$? Commented Jan 29, 2020 at 13:08
• "The L cannot be equal to both 0 and 1"
– Tim
Commented Jan 29, 2020 at 13:09
• I think that it is quite trivial that no number can be simultaneously equal to $0$ and to $1$. Do you disagree? Commented Jan 29, 2020 at 13:11
• Actually, I wrote $(-\delta,\delta)$, not $(-\delta,-\delta)$. Anyway, $0\in(-\delta,\delta)$ and $0$ is rational, right?! More generally, you can use the fact that the rationals are dense: every interval $(a,b)$ of real numbers contains rational numbers. Commented Jan 29, 2020 at 13:20

If possible let $$f(x) \to L \neq 0$$ so $$|f(x)-L| <\frac {|L|} 2$$ for $$|x| <\delta$$ for some $$\delta$$. Choosing $$x$$ rational with $$|x| <\delta$$ we get $$|L| <\frac {|L|} 2$$. This is a contradiction so we must have $$L=0$$. Now there is a $$\delta$$ such that$$|f(x) |<\frac 1 2$$ for $$|x| <\delta$$ and we get contradiction again by taking $$x$$ irrational with $$|x| <\delta$$.