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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.

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  • $\begingroup$ 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
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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 $$\lim_{x\to 0} f(x) = 0$$

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