I know that $\displaystyle \mathbb{Z}/(6)^i\mathbb{Z}\cong \mathbb{Z}/(2)^i\mathbb{Z}\times \mathbb{Z}/(3)^i\mathbb{Z}$ for any $\displaystyle i$.

But $\displaystyle \varprojlim_{i\in I} \mathbb{Z}/(6)^i\mathbb{Z}\cong \varprojlim_{i\in I} \mathbb{Z}/(2)^i\mathbb{Z}\times \varprojlim_{i\in I} \mathbb{Z}/(3)^i\mathbb{Z}??$

Please give an explanation if it's right.