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Given a function

$$F(x)= \begin{cases} x^2 & \text{when }x \in \mathbb Q \\3x & \text{when }x \in\mathbb Q^c \end{cases}$$

Show that $F$ is continuous or not on $x=3$ with $\epsilon-\delta$.

I tried to deal with problems just like doing on Dirichlet functions. Mistakenly or not, I couldn't. Can anyone help me, please?

user276702
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    What did you try? –  Oct 04 '15 at 06:12
  • I supposed epsilon is 1/2. And tried to find contradiction. And no conclusion still. – user276702 Oct 04 '15 at 06:34
  • What is your guess? Is $f$ continuous at $x = 3$? –  Oct 04 '15 at 06:37
  • I guess it is not continuous at x=3. And with (limit of f at x=3)=L and epsilon=1/2 , i tried to find contradiction. – user276702 Oct 04 '15 at 06:41
  • However, if you plug in $x=3$ into both $3x$ and $x^2$, they are both $9$. It is an indication that the function $f$ is continuous there. –  Oct 04 '15 at 06:43
  • Both x^2 and 3x are continuous at 3, and both assume the same value 9. So F is clearly continuous at 3. – Vim Oct 04 '15 at 07:07

3 Answers3

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Let $\epsilon > 0$ , be a positive number

To find a $\delta$ such that $|x-3|<\delta$ implies $|f(x)-f(3)|<\epsilon$.

Case-I if $x\in \mathbb Q$

Then $|f(x)-f(3)|=|x^2-9|=|(x-3)(x+3)|$

Therefore we choose our $\delta$ to be $<1$, then $|x-3|<1$ implies $-1<x-3<1$ , adding 6 on both sides we get $ x+3<7$.

$|f(x)-f(3)|=|x^2-9|=|(x-3)(x+3)|<|x-3|\cdot 7$

Therefore we can choose, $\delta $ to be $\dfrac{\epsilon}{7}$ for the case-I

Case-II , $ x $ is irrational number,

$|f(x)-f(3)|=|3x-9|=|3(x-3)|$

Therefore we can choose, $\delta $ to be $\dfrac{\epsilon}{3}$ for the case-II

So, for all real $x$ we can choose $$\delta= min\{1,\frac{\epsilon}{7}, \frac{\epsilon}{3} \}$$ which ends the proof of $f$ being continuous at 3.

Note: 1) For a better and clear understanding of manipulation of $\delta $ you can refer to this beautifully written answer which got 27 upvotes-How to show that $f(x)=x^2$ is continuous at $x=1$?

Community
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Babai
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    Oh! Thank you So much for detailed solution and informing another good things. I appreciate it. – user276702 Oct 04 '15 at 06:55
  • One more observation, $\delta= min{1,\frac{\epsilon}{7}, \frac{\epsilon}{3} }$ is same as $\delta= min{1,\frac{\epsilon}{7} }$ – Babai Oct 04 '15 at 06:57
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Hint: Note that $x^{2} \geq 3x$ for all $x \geq 3$ and that $x^{2} \leq 3x$ for all $0 \leq x \leq 3$; now $ |F(3+h) - F(3)| \leq (3+h)^{2} - 9 = 6h+h^{2} \to 0 $ as $h \to 0+$ and $ |F(3+h) - F(3)| \leq |(3+h)^{2} - 9| = |6h+h^{2}| \to 0 $ as $h \to 0-$, so $|F(3+h) - F(3)| \to 0$ as $h \to 0$.

Yes
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let $\epsilon>0$,let $\delta = min(\frac{\epsilon}{7},1)$. let $x$ satisfy $|x-3|<\delta$, if $x \in Q $, then we have:

$|F(x) - F(3)| = |x^2-9| = |x-3||x+3| \leq 7|x-3| < 7\delta \leq \epsilon$

if $x$ not in $Q$ we have:

$|F(x) - F(3)| = |3x-9| = 3|x-3| < 3\delta \leq 3 \frac{\epsilon}{7} < \epsilon$

so, for all x that satisfy $|x-3|<\delta$ we have $|F(x) - F(3)|<\epsilon$, from here we see that $F(x)$ is continuous on x=3.

D. A.
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