Hfrhyu

Define $g: mathbbR to mathbbR$ by $$g(x) =begincases x^2 & textif x text is rational, \\ x^4 & textif x text is irrational. endcases$$ Prove that $g$ is discontinuous at $x = 2$.

Solution Attempt:

Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_xto2g(x)=4$;

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

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

$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac14 implies x < 2$ Contradiction

Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
$$-sqrt[4]5 < x < sqrt[4]5.$$
If you had chosen instead an irrational $x$ such that
$$2 - mindelta, 2 - sqrt[4]5 < x < 2,$$
then you'd have $x > 2 - (2 - sqrt[4]5) = sqrt[4]5$, 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 $lim_ntoinftyg(a_n)=2^2$ and $lim_ntoinftyg(b_n)=2^4$.

Then we can say $lim_xto2^+g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.

Put $a_n = 2 + frac1n$ and $b_n=2+fracpin$.

An option?

Sequential definition of continuity .

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

$lim_n rightarrow inftyx_n=2.$

$lim_n rightarrow inftyg(x_n)=$

$lim_n rightarrow infty 2^4(4n)^(1/n)=$

$lim_n rightarrow infty2^4 cdot 4^(1/n) cdot n^(1/n)=$

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

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

$$lefta_n:=2pifrac npi n+1right_ninBbb N$$

Observe the above sequence is irrational:

$$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,textso;;lim_ntoinftyg(a_n)=lim_ntoinfty16pi^4fracn^4(npi+1)^4=16pi^4frac1pi^4=16$$

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