I am only allowed to use the fact that a certain sequence {${x_i}$} that is not equivalent to the zero sequence is cauchy to prove that the reciprocal {$\frac{1}{x_i}$} is also cauchy. i.e. we know that $\forall \epsilon > 0$, $\exists\,N_{\epsilon} \in \mathbb N$ s.t. if $i, j > N_{\epsilon}$, then $|x_i - x_j| < \epsilon$ Define $\epsilon ^{,}$ s.t $\epsilon ^{,}$ = min{$\epsilon, \epsilon |x_i x_j|$}. So if we have $i, j > N_{\epsilon^{,}}$
$$|x_i - x_j| < \epsilon ^{,}$$
$$\implies |x_i -x_j| < \epsilon |x_i x_j|$$
$$\implies \frac{|x_i -x_j|}{|x_i x_j|} < \epsilon $$
$$\implies |\frac{x_i -x_j}{x_i x_j}| < \epsilon $$
$$\implies |\frac{1}{x_j} - \frac{1}{x_i}| < \epsilon $$
which implies that the reciprocal function is cauchy. Is my proof rigorous enough? I'm especially skeptical of this step where I chose my $\epsilon ^{,}$. It seems antithetical to the idea of limits.