Determine whether $F(x)$ is uniformly continuous or not in Real numbers

Question

$$f(x)= \begin{cases} \frac{3}{4}x + 1, & x\in \mathbb Q\\ 3 + \frac{1}{2}x, & x\notin \mathbb Q\ \end{cases}$$

I have proved with the help of sequence method that f is not continuous in $\mathbb R$ except for $x$ equal to 8. But now I need to prove that f is continuous at $x=8$ from the definition of epsilon and delta. I found delta to be $4/3(\epsilon - 1)$ but then for $0<\epsilon<1$ it will not work. How to show continuity at $x=8$ then?

Answer 1

Note that $f(8)=7$. Calim: $\delta =4\epsilon /3$ suffices: $|\frac {3x} 4+1-7|=\frac {3|x-8]} 4<\epsilon$ if $|x-8| <4\epsilon /3$. Also $|3+\frac {x} 2-7|=\frac {|x-8]}2 <\epsilon$ if $|x-8| <4\epsilon /3$.

Answer 2

Since $8\in\Bbb Q$, $f(8)=7$. So, if $\varepsilon>0$, you want to prove that there is some $\delta>0$ such that$$|x-8|<\delta\implies\bigl|f(x)-7\bigr|<\varepsilon.$$But$$\bigl|f(x)-7\bigr|=\begin{cases}\left|\frac34x-6\right|=\frac34|x-8|&\text{ if }x\in\Bbb Q\\\left|\frac x2-4\right|=\frac12|x-8|&\text{ if }x\notin\Bbb Q.\end{cases}$$So, you always have $\bigl|f(x)-7\bigr|\leqslant\frac34|x-8|$ and therefore $\delta=\frac43\varepsilon$ will work for every $\varepsilon$.