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Calculus for Scientists and Engineers: Early Transcendental

Proving that $\lim _{x \rightarrow a} f(x) \neq L$ Use the following definition for the nonexistence of a limit. Assume $f$ is defined for all values of $x$ near a, except possibly at a. We say that $\lim _{x \rightarrow a} f(x) \neq L$ iffor some $\varepsilon>0$ there is no value of $\delta>0$ satisfying the condition $$ |f(x)-L|<\varepsilon \text { whenever } 0<|x-a|<\delta $$

Let

$$

f(x)=\left\{\begin{array}{ll}

0 & \text { if } x \text { is rational } \\

1 & \text { if } x \text { is irrational. }

\end{array}\right.

$$

Prove that $\lim _{x \rightarrow a} f(x)$ does not exist for any value of $a$. (Hint:

Assume $\left.\lim _{x \rightarrow a} f(x)=L \text { for some values of } a \text { and } L \text { and let } \varepsilon=\frac{1}{2} .\right)$