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**dwsmith** Define $\displaystyle f:\mathbb{R}\to\mathbb{R}$ by $\displaystyle f(x)=8x, \ \text{if} \ x\in\mathbb{Q}$ and $\displaystyle f(x)=2x^2+8$ if x is irrational. Prove from the definition that f is discontinuous at 1.

Let $\displaystyle \epsilon >0$. Then there exist $\displaystyle \delta_1, \ \delta_2>0$.

$\displaystyle |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}$.

Let $\displaystyle \delta_1=\frac{\epsilon}{8}$.

$\displaystyle |2x^2+8-10|=2|x^2-6|<\epsilon$.

Does it suffice to say that $\displaystyle |x^2-6|$ can't be manipulated to some form of $\displaystyle |x-1|$ now?