$ \lim_{n\rightarrow \infty}[a_n-a]=0 $ implies $ \lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]=0 $? Suppose that 
$$
\lim_{n\rightarrow \infty}[a_n-a]=0
$$
Does this imply 
$$
\lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]=0
$$
? Does the answer change if in place of $a$ we have another sequence $b_n$?
Yes. We have that $\lim_{n\to\infty}[a_nb_n]=\lim_{n\to\infty}a_n\lim_{n\to\infty}b_n$ if the limits on the right hand side exist. Hence,
$$
\lim_{n\to\infty}\frac1n[a_n-a]=\lim_{n\to\infty}\frac1n\lim_{n\to\infty}[a_n-a]=0\cdot0=0.
$$
If we replace $a$ with some sequence, the limit might not be $0$ anymore (take, for example, $a_n=1/n$ and instead of $a=0$ take $b_n=n$).
For constant $a$:

$$
\lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]
$$

$$
=\lim_{n\rightarrow \infty}\frac{1}{n}\lim_{n\rightarrow \infty}[a_n-a]=(0)(0)
$$

For sequence $b_n$:

$$
\lim_{n\rightarrow \infty}\frac{1}{n}[a_n-b_n]
$$

$$
=\lim_{n\rightarrow \infty}\frac{1}{n}\lim_{n\rightarrow \infty}[a_n-b_n]=(0)(0)
$$



Remark: Wherever they exist, limits of sequences behave like limits of functions, i.e. We can split of limits by products into products of limits. This is because a sequence is a function whose domain is countable.