Prove that a function is not continuous at a point using the proof of negation I am given a piecewise function
$$f(x) = \begin{cases} \cfrac{1}{x} & x \neq 0 \\ -10 & x = 0\end{cases},$$
and I am trying to prove that the function is discontinuous at $x = 0$ by the proof of negation. So far, I am thinking about proving that there exists an $\epsilon > 0$ such that for all $\delta > 0$, $| x - 0 | = | x | < \delta$ but $| f(x) - f(0) | = \left| \frac{1}{x} + 10 \right| \ge \epsilon$, but I am kinda stuck here. I am not sure of my negation of the delta-epsilon definition is correct.
By some exploratory analysis, I found that $\frac{1}{| x |} > \delta$, and by the triangle inequality, $\left| \frac{1}{x} + 10 \right| \ge \frac{1}{| x |} + 10 > \delta + 10$. Then, should I choose $\epsilon = \delta - 10$, so that $\left| \frac{1}{x} + 10 \right| > \epsilon - 10 + 10 = \epsilon$? My concepts are so messed up. Please help me if you can. Thank you so much.

The logical negation of
$$ \forall \epsilon>0\colon\exists \delta>0\colon\forall x\in(-\delta,\delta)\colon|f(x)-f(0)|<\epsilon$$
is (by "inverting" every quantor and the order relation symbol)
$$ \exists \epsilon>0\colon\forall \delta>0\colon\exists x\in(-\delta,\delta)\colon|f(x)-f(0)|\ge\epsilon$$
To prove this then, picking $\epsilon=9$ and later $x=\frac12\delta$ will help.
