# Proving that $g(x)$ defined as $x^2$ on the rationals and $x^4$ on irrationals, is discontinuous at $2$

Define $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) =\begin{cases} x^2 & \text{if } x \text{ is rational}, \\\\ x^4 & \text{if } x \text{ is irrational}. \end{cases}$$ Prove that $$g$$ is discontinuous at $$x = 2$$.

Solution Attempt:

Assume instead that $$g$$ is continuous at 2. Then since $$g(2)= 4$$, $$\;\;\displaystyle \lim_{x\to2}g(x)=4$$;

so taking $$\epsilon=1$$, there is a $$\delta>0$$ such that if $$0<|x-2|<\delta$$, then $$\big|g(x)-4\big|<1$$.

Therefore if $$x$$ is irrational and $$2-\delta, then $$\big|x^4-4\big|<1$$

$$-1 3 \implies x = 3^\frac{1}{4} \implies x < 2$$ Contradiction

• Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0) Apr 7, 2019 at 6:03
• $x<2$ is not a contradiction. What you should note is that you can choose $\delta$ arbitrarily small, so choose $\delta$ such that $3^{1/4}<2-\delta$. Then you get a contradiction. Apr 7, 2019 at 6:07

Rather than $$x^4 > 3$$, instead conclude that $$x^4 < 5$$, or equivalently, $$-\sqrt{5} < x < \sqrt{5}.$$ If you had chosen instead an irrational $$x$$ such that $$2 - \min\{\delta, 2 - \sqrt{5}\} < x < 2,$$ then you'd have $$x > 2 - (2 - \sqrt{5}) = \sqrt{5}$$, a contradiction.

I think a simpe way is to construct two sequences $$a_n$$ and $$b_n$$, where $$a_n$$ are rational and $$b_n$$ are irrational, such that $$\lim_{n\to\infty}g(a_n)=2^2$$ and $$\lim_{n\to\infty}g(b_n)=2^4$$.

Then we can say $$\lim_{x\to2^+}g(x)$$ does not exist. Hence $$g$$ is discontinuous at $$x=2$$.

Put $$a_n = 2 + \frac{1}{n}$$ and $$b_n=2+\frac{\pi}{n}$$.

An option?

Sequential definition of continuity .

$$x_n=2(4n)^{(1/(4n))}.$$

$$\lim_{n \rightarrow \infty}x_n=2.$$

$$\lim_{n \rightarrow \infty}g(x_n)=$$

$$\lim_{n \rightarrow \infty} 2^4(4n)^{(1/n)}=$$

$$\lim_{n \rightarrow \infty}2^4 \cdot 4^{(1/n)} \cdot n^{(1/n)}=$$

$$=2^4 \not =g(2)$$.

You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take

$$\left\{a_n:=2\pi\frac n{\pi n+1}\right\}_{n\in\Bbb N}$$

Observe the above sequence is irrational:

$$a_n\notin\Bbb Q\implies g(a_n)=a_n^4\,,\,\,\text{so}\;\;\lim_{n\to\infty}g(a_n)=\lim_{n\to\infty}16\pi^4\frac{n^4}{(n\pi+1)^4}=16\pi^4\frac1{\pi^4}=16$$

yet $$\;g(2)=2^2=4\;$$ , so the function is not continuous at $$\;2\;$$ ...