Natural transformations preserve exactness. Is there a quick way to see that if F,G are both functors (say between abelian categories or $R-Mod$ to Ab) and $F \cong G$, then F exact implies G exact?

Note that a functor is left exact if and only if it preserves finite limits. Now, assume $F$ is left exact and let $D$ be a finite diagram. Then we have $$\operatorname{Hom}(X,\lim GD)\cong \operatorname{Hom}(X,\lim FD)\cong \operatorname{Hom}(X,F\lim D)\cong \operatorname{Hom}(X,G\lim D)$$ where all isomorphisms are due to $F\cong G$. Thus, $G$ preserves finite limits and therefore is left exact. Right exactness follows dually.
