This question is stuffing me over.

Let $f$ be defined by $$f(x) = \begin{cases} x & \text{if } x \in \mathbb Q \\ -x & \text{if } x \notin \mathbb Q \end{cases}$$

Use the definition of a limit to show that $\lim_{x\to 0} f(x)$ exists.

I get how for a limit to exist we need to show that for every $\epsilon$ that is greater than $0$ there exists a $\delta$ that greater than $0$, such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$, but I don't know how to actually do it.

  • $\begingroup$ Please make it latex format otherwise you'll get a lot of down votes soon. $\endgroup$ – user175968 Mar 21 '16 at 20:51
  • $\begingroup$ sorry I don't know how to use latex format $\endgroup$ – Arthur King Lee Mar 21 '16 at 20:58
  • $\begingroup$ sorry I don't know how to use latex format and my computer doesn't seem to like the latex format $\endgroup$ – Arthur King Lee Mar 21 '16 at 20:59
  • $\begingroup$ Is $\;f\;$ a complex function? Or is it defined in any structure containing the real numbers? Otherwise its definition makes no sense, I believe. $\endgroup$ – DonAntonio Mar 21 '16 at 21:12
  • $\begingroup$ Yes you are right its Q instead of R $\endgroup$ – Arthur King Lee Mar 21 '16 at 21:30

First of all, what would the limit be? If we just look at the rational part $f(x) = x$, whose limit as $x \to 0$ is just $0$, as I hope you already know. If $f$ is going to converge at all, it has to converge to $0$. So in the definition you gave, $L = 0$. So $|f(x) - L| = |f(x)|$. Now, we are also taking the limit at $0$, so $a = 0$ as well. Thus $|x - a| = |x|$. With these two values, the definition becomes:

For every $\epsilon > 0$, there is a $\delta > 0$ such that if $|x| < \delta$, then $|f(x)| < \epsilon$.

That means: if we are given some value $\epsilon > 0$, then we can find some value $\delta > 0$ (which can depend on $\epsilon$) so that we can show $|x| < \delta$ implies that $|f(x)| < \epsilon$.

For this problem, it is easy to do. Let's look at $|f(x)|$. If $x \in \Bbb Q$, then $f(x) = x$, so $|f(x)| = |x|$. If $x \notin \Bbb Q$, then $f(x) = -x$, and therefore $|f(x)| = |-x| = |x|$. Thus for every value of $x$, $|f(x)| = |x|$.

So now, the statement that we have to prove becomes:

For every $\epsilon > 0$, there is a $\delta > 0$ such that if $|x| < \delta$, then $|x| < \epsilon$.

Obviously, we should just choose $\delta = \epsilon$.

That is the planning for the proof. Not the actual proof, which must start with what we are given, and deduce the conclusion from it. It goes like this:

Proof: Let $\epsilon > 0$ be given, and set $\delta = \epsilon$. If $|x| < \delta = \epsilon$, then $|f(x)| = |\pm x| = |x| < \epsilon$. Hence $$\lim_{x\to 0} f(x) = 0$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.