For positive definite matrix, if we increase the dimension to the infinity, is it true that the largest eigenvalue stays bounded from above? In other words does the following limit exists:

$$\lim_{n\rightarrow \infty }\frac{\lambda_{max}}{n}=C<\infty,$$

where $\lambda_{max}$ is the largest eigenvalue of $n\times n$ symmetric positive definite matrix.

For every $n$ the eigenvalues are bounded from above, but is it true in the limit? It is not exactly infinite matrix per se, but finite matrices increasing in its size.

This question is related to my previous question, but is of independent interest in general and no information from my other question is necessary.

UPDATE

We have a sequence of matrices, but the mechanism of the sequence formation is not known. Only known is that it has to be kept positive definite and symmetric. Therefore we have a sequence of matrices such that each new $n^{th}$ matrix contains the previous matrix plus new added row and column. Hence we have a sequence of positive definite matrices, increasing in size with previous matrix being a sub-matrix. In addition, all values of the matrix has to be bounded from above (and below), no matter how large the matrix gets.

Probably this is way too general a problem to be able to infer anything usefull.