$\delta{-}\epsilon$ piecewise discontinuity proof by contradiction

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Let $f(x)$ be given by

$$f(x) = \left\{\begin{array}{ll} 1 & : x \ge 0\\ 0 & : x<0 \end{array} \right.$$

Show that there is no L for which $\lim_{x\to0} f(x) = L$

Hint: You’ll want to assume that $\lim_{x\to0} f(x) = L$ and, from this assertion, derive a contradiction.

You have a number of options here; one vector of attack would be showing that if $\lim_{x\to0} f(x) = L$ we necessarily have $L > 1/2$ and $L < 1/2$.

Another would be to show that $\lim_{x\to0} f(x) = L$ implies that $1 < 1$ which is obviously absurd.

So I've been struggling with this for a few hours and have made no progress. My teacher said something about setting $\epsilon = 1/2$ but I have no idea how to incorporate this.

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Use the definition of continuity, then show that there is no $\delta$ that works for $\epsilon$ = 1/2

Also, it is helpful to write down and understand the negation of the definition of continuity.