Let $f(x) = \frac{1}{x}$ be a function to and from the reals. Is $f$ bounded?
I want to prove using the definition. Let $M > 0$ be given. Then there exists a $\delta > 0$ such that:
If $0<|x-0| < \delta$ implies $|f(x)|=|\frac{1}{x}|=\frac{1}{|x|}>\frac{1}{\delta} = M$ whenever $\delta = \frac{1}{M}$
So is it correct to pick $\delta = \frac{1}{M}$ and I have proved that as $x$ tends to $0$ which, $f(x)$ tends to positive infinity, so it is unbounded.
On
The formulas that you've written actually prove something stronger, namely that $$ \lim_{x \to 0}\; \biggl\lvert \frac1x \biggr\rvert = \infty. $$
To establish that $f(x) = \frac1x$ is unbounded, we have to show that it is either unbounded from above or from below. The following applies to the upper bound, but everything works mutatis mutandis to show that for this function there's no lower bound either.
For each potential upper bound $M$, we have to demonstrate that there's some $x$ in the domain of $f$ such that $f(x) > M$. Your formulas show that it suffices to choose any $x \in \smash{\bigl(0, \frac1M \bigr)}$, but such a universal quantifier isn't strictly necessary here. In other words, we just need an example point $x$ rather than all possible $x$ in a neighborhood of $0$.
If $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ was bounded there would exist a constant $M >0$ such that $$ |f(x)| < M, \quad \forall x\ne 0. $$
However, this generates a contradiction since $|f(1/M)| > M$.