Show that Let $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ be defined by $f(x) = \frac{1}{x}$. Show $$\lim_{x \to 0}f(x)$$ doesn't exist.
I don't think I've ever seen a rigorous proof of this. I'm trying to show rigorously that this limit doesn't exist.
My Attempted Proof: Observe that $0$ is a limit point of $\mathbb{R} \setminus \{0\}$ so it makes sense to talk about the limit of $f(x)$ as $x \to 0$. Suppose that $$\lim_{x \to 0} f(x) = q \in \mathbb{R}.$$ Then for all $\epsilon > 0$ there exists a $\delta > 0$ such that $|x- 0|< \delta \implies \left|\frac{1}{x} - q\right| < \epsilon$.
Now this is where I got stuck, I want to show that a contradiction arises from the above which will allow me to conclude that no such $q \in \mathbb{R}$ exists. The only way to do this (I think) would be to find an $\epsilon > 0$ such that for all $\delta > 0$ we have $|x| < \delta$ and $\left|\frac{1}{x} - q\right| > \epsilon$. I'm having trouble finding such an $\epsilon$. The only candidate I can think of is $\epsilon = \frac{1}{|q|}$ but I'm still not sure how two conclude that $\left|\frac{1}{x} - q\right| > \frac{1}{|q|}$.