inverse limit and graded functor commute?

Question

I am trying to understand a proof from Okounkov-Olshanski, Shifted Schur functions's paper. Exactly Proposition 1.5. which says

PROPOSITION 1.5. The graded algebra $gr(\Lambda^*)$ corresponding to the filtered algebra $\Lambda^*$ is canonically isomorphic to the algebra $\Lambda$.

where $\Lambda^*$ is the algebra of shifted symmetric functions and $\Lambda$ is the algebra of symmetric functions.

In the proof, they state that $\varprojlim gr(\Lambda^*) = \Lambda^*$. I guess that this is by saying that $gr(\varprojlim \Lambda_n^*) = gr(\Lambda^*)$ since $\Lambda^*=\varprojlim \Lambda_n^*$ (and taking their respective graded algebras) and then saying that $gr(\varprojlim \Lambda_n^*)=\varprojlim gr(\Lambda_n^*)$.

Note that $\Lambda^*=\varprojlim \Lambda_n^*$ is a projective limit in the category of filtered algebras.

QUESTION:

1. Do inverse limit and graded functor commute?

2. How can we see that $\varprojlim gr(\Lambda^*) = \Lambda^*$?

3. Is there any reference or result that states $gr(\varprojlim \Lambda_n^*)=\varprojlim gr(\Lambda_n^*)$? (i.e. the $gr$ functor and the inverse limit commute?)

MO QUESTION HERE.