What is the formal proof that the reciprocal of a number approaches 0 as the number increases without bound? In Calculus textbooks it seems assumed that
$$ \lim_{n\to\infty} \frac{1}{n} = 0 ,$$
meaning that "the limit of the reciprocal of $n$ as $n$ increases without bound is zero".
This makes intuitive sense: if $a/b$ gives the (potentially fractional) number of times $b$ fits into $a$, then as $b$ increases without bound, $a/b$ should get increasingly small.
What would be the formal proof that $$\lim_{n\to\infty} \frac{1}{n} = 0?$$

More generally, this is usually expressed as
$$ \lim_{n\to\infty} \frac{a}{n} = 0 ,$$
though if $a$ does not contain $n$ it gives the same result,
$$ \lim_{n\to\infty} \frac{a}{n} = a \lim_{n\to\infty} \frac{1}{n} = a(0) = 0 .$$
 A: Consider the set $A = \{1/n\mid n\in\mathbb{N}_{>0}\}$.
Clearly, it is bounded below by $0$ and it is not empty. Consequently, it admits an infimum $L\geq 0$.
Suppose that $L > 0$. Hence $2L > L$, which means there exists an $n_{0}$ such that $2L > 1/n_{0}$.
In other words, we have just proven that $L > 1/2n_{0}$, which contradicts our assumption.
Thus we can conclude that $\inf(A) = 0$.
Now we are ready to prove the proposed claim. More precisely, we want to prove the following proposition:
\begin{align*}
(\forall\varepsilon > 0)(\exists n_{0}\in\mathbb{N}_{>0})\:\text{s.t.}\left(n\geq n_{0}\Rightarrow \frac{1}{n} < \varepsilon\right)
\end{align*}
Indeed, this is the case.
Let us consider an arbitrary positive real number $\varepsilon > 0$.
Given that $\varepsilon > \inf(A)$, there exists an $n_{0}$ such that $\varepsilon > 1/n_{0}$.
Consequently, for every $\varepsilon  > 0$, there corresponds an $n_{0}\in\mathbb{N}_{>0}$ such that
\begin{align*}
n\geq n_{0} \Rightarrow \frac{1}{n} \leq \frac{1}{n_{0}} < \varepsilon
\end{align*}
Hopefully this helps!
A: To show that $\lim_{n\to\infty} \frac 1n=0$, we need to show that, for every $\epsilon$, there exists a threshold $n_0$ for which, if $n>n_0$, then $\left|\frac 1n\right|\leq \epsilon.$
Take any $n_0$ greater than $1/\epsilon$. Then, for any $n>n_0$,
$$0\leq \frac 1n<\frac{1}{n_0}<\epsilon,$$
and we are done.
(This last chain of inequalities corresponds to $1/\epsilon<n_0<n$.)
A: We make the concept of "approach" in limits rigorous through epsilon-delta proofs, which you can look up.
Here is the proof for your limit: limit as $x$ approaches infinity of $\frac{1}{x}$
A: Hint
You are actually referring to the limit of a sequence
$$1, (1/2), (1/3), \cdots $$
rather than the limit of a function.
Given $\epsilon > 0,~$ take $M \in \mathbb{Z^+}$ such that $$\frac{1}{\epsilon} < M.$$
