I am doing a question (q-39 from ch 2.4- Stewart's Calculus) that asks me to prove that for the function
$$f(x)=\begin{cases}
0, & \text{if } x \text{ is rational} \\
1, & \text{if } x \text{ is irrational}
\end{cases}$$
the $\lim_{x\to 0}f(x)$ does not exist.
I am stuck because I have not dealt with the limits of conditional functions using delta epsilon. I understand that the start of the proof would be to assume $|f(x)-L|<\epsilon$. However the solution set suggests that it follows that $\epsilon=1/2$. How do I get $\epsilon=1/2$ from $|f(x)-L|<\epsilon$?
Best Answer
Ok, not $ε-δ$ but $\lim_{x\to0}\sup f(x)=1$, $\lim_{x\to0}\inf f(x)=0$.
To prove it with $ε-δ$, see here:
Still, quoting from the above link:
So $ε=1/2$ is just a choice. Any choice of $0<ε<1$ would do, but $ε=1/2$ is simple.