How to prove that the limit of a piecewise defined function does not exist?

Question

If we define $$ f(x)=\begin{cases} x&x\geq0\\ -1&x<0 \end{cases} $$ To prove that $\lim_{x\to0}f(x)$ does not exist, what am I required to do?

I already know that if $\lim_{x\to a^+}f(x)\not=\lim_{x\to a^-}f(x)$, then $\lim_{x\to a}f(x)$ does not exist. Is stating this theorem the proof of the above question or in order to prove it I somehow need to use precise definition of the limit?

Answer 1

It's enough, since you have $$\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0\text{ and}\lim_{x\to0^-}f(x)=\lim_{x\to0^-}-1=-1$$ Therefore the limit doesn't exist.

Also both $\lim\limits_{x\to0}x=0$ and $\lim\limits_{x\to0}-1=-1$ are immediately obvious from the definition.