Suppose $K$ is fixed and consider a matrix $X$ that is $n\times K$ and has full column rank. Assume that we know $$ \frac{1}{n}X'X\to M\text{ as } n\to\infty.\tag{i} $$ That is, as $n$ becomes larger, we add more rows to $X$ in such a way that $X$ still has full column rank and (i) holds. It is given that $M$ is a fixed positive definite matrix with all of its eigenvalues in $(b,B)$ with some $0<b<B$.
Now, let $X_i$ be the $K\times 1$ vector such that $X'_i$ is the $i$-th row of $X$. Does it follow that
$$ \lim_{n\to\infty}\frac{1}{n}\max_{1\leq i\leq n}|X_i|^2=0?\tag{ii} $$
I tried looking at $$ \frac{1}{n}X'X=\frac{1}{n}\begin{pmatrix}X_1&\cdots& X_n\end{pmatrix}\begin{pmatrix}X'_1\\\cdots\\X'_n\end{pmatrix}=\frac{1}{n}\sum_{i=1}^n X_iX'_i\to M.\tag{iii} $$ It seems intuitive that each summand in (iii) has to be "small" and perhaps that will eventually leads me to (ii) but I can't produce a decent argument. Can someone help please?
Edit: I also looked at $$ \text{Tr}\left(\frac{1}{n}X'X\right)=\frac{1}{n}\sum_i\text{Tr}(X_iX_i')=\frac{1}{n}\sum_i |X_i|^2\to\text{Tr}(M)\tag{iv} $$ but I reached a deadend here too.
As you observe, $$ \text{Tr}\left(\frac{1}{n}X'X\right)=\frac{1}{n}\sum_i\text{Tr}(X_iX_i')=\frac{1}{n}\sum_i\text{Tr}(X_i'X_i)=\frac{1}{n}\sum_i|X_i|^2. $$ So $\frac{1}{n}X'X\to M$ implies $\frac{1}{n}\sum_i|X_i|^2\to\text{Tr}{M}$ with the latter being a positive number in $(Kb, KB)$. Then, the claim $$ \lim_{n\to\infty}\left(\max_{1\leq i\leq n}|X_i|\right)^2=0 $$ follows from the results discussed here.