Suppose $H$ is an $n\times n$ symmetric positive definite matrix, $M_k$ is a sequence of $n \times n$ matrix (**not necessarily symmetric**) such that $M_k \to O$ where $O$ is the zero matrix. Let $\lambda_i(H),i=1,...,n$ denote the operator that gets the $i$th largest eigenvalue of $H$ in absolute value. My question is, is it true that $\mathop {\lim }\limits_{k \to \infty } \lambda_i (H + {M_k}) = \lambda_i (H),i=1,...,n$?

If this is not true for every $i$, is it true for $i=1$ (largest eigenvalue) and $i=n$ (smallest eigenvalue)? (**My application only needs this one to hold**)

Any explanation, counterexample or reference is helpful. Thanks!