Consider three real numbers, $a,b,c\in \mathbb{R}$.
What I want to prove is $ab-c\neq 0$.
For now, I derived the following condition: $\forall \delta \in \mathbb{R}\backslash${$0,1$}, $a(b/\delta)-c\neq 0.$
And, I considered a proof: $\textrm{Fix a sequence } \delta_n\subset \mathbb{R}\backslash${$0,1$}$\textrm{ such that }\lim_{n\rightarrow\infty}\delta_n=1, \textrm{ then } (a(b/\delta_n)-c)\neq 0 \textrm{ and }\lim_{n\rightarrow \infty}(a(b/\delta_n)-c)\neq \lim_{n\rightarrow \infty}0,\textrm{ thus }ab-c\neq 0.$
However, I think, $(a(b/\delta_n)-c)\neq 0$ $\Rightarrow$ $\lim_{n\rightarrow \infty}(a(b/\delta_n)-c)\neq \lim_{n\rightarrow \infty}0$ does not hold without additional assumptions.
So, does $(a(b/\delta_n)-c)\neq 0$ $\Rightarrow$ $\lim_{n\rightarrow \infty}(a(b/\delta_n)-c)\neq \lim_{n\rightarrow \infty}0$ hold? If doesn't, what assumption is needed?