Proposition. Assume $x$ and $a$ are elements of $\mathbb{R}_{\neq 0}$. Then if $x$ is near to $a$, we can deduce that $x^{-1}$ is near to $(a+a^{-1}) - x.$
Proof. Since the function $x^{-1} - (a+a^{-1}) + x$ vanishes at $x = a$ and since it's continuous when $x \neq 0$ and $a \neq 0$, we're done.
Remark. I derived the formula like so: If $x = a$, then $x - a = 0$, thus $(x - a)(x-a^{-1}) = 0$. Now expand the brackets, divide through by $x$, and get the resulting $x^{-1}$ by itself on one side of the equation.
Example. Observe that $1.01$ is near to $1$. Thus $1.01^{-1}$ is near to $(1+1)-1.01$. So the reciprocal of $1.01$ is approximately $0.99$.
Question. Is there an explicit bound we can find? I.e. can we find $f$ such that $$|x^{-1} - (a+a^{-1}) + x| < f(|x - a|)$$
Remark. It might be helpful to look at this more symmetrically as follows:
$$|(x+x^{-1}) - (a+a^{-1})| < f(|x - a|)$$
We cannot. In other words, $g(x) = x + x^{-1}$ fails to be uniformly continuous over $\Bbb R_{\neq 0}$.
For a fixed $\delta > 0$ and any $x$, we can take $a = x + \delta$. We find that $\lim_{x \to 0^+} |g(x) - g(a)| = \infty$, so that there is no value for $f(|x-a|) = f(\delta)$ for which $|g(x) - g(a)| < f(|x-a|)$ holds for all $x \in \Bbb R_{\neq 0}$.
For $a,x$ in a closed subinterval $[p,q]$ of $\Bbb R_{\neq 0}$, it suffices by the mean value theorem to take $f(|a - x|) = \max_{t \in [p,q]}|g'(t)|\cdot |a - x|$.