If $V$ is a complex inner product space, regard $\textrm{End}_{\mathbb{C}}(V)$ as a Banach space with the operator norm. For $T \in \textrm{End}_{\mathbb{C}}(V)$ as I understand, the singularvalues $\sigma_1(T),\cdots,\sigma_n(T)$ are continuous functions of $T$ (as their are the square roots of the eigenvalues of $TT^*$). On the other hand the set of diagonalizable operators in $\textrm{End}_{\mathbb{C}}(V)$ is dense in $\textrm{End}_{\mathbb{C}}(V)$, and for diagonalizable operators the singular values are the absolute of eigenvalues.
Now for any $T \in \textrm{End}_{\mathbb{C}}(V)$, if we consider a sequence of diagonalizable transformations $\{T_k\} \rightarrow T$, by the above argument $\sigma_i(T_k) \rightarrow \sigma_i(T), i=1,\cdots,n$. Thus it follows that the singular values of $T$ should also be the absolute of its eigenvaules, which is obviously false. Something is wrong here, but I can't figure out what is incorrect in this argument. Can anyone help me know what I am misunderstanding here? Thank you.
As Matsmir commented, normality of the operator is necessary for $\sigma_i = |\lambda_i|$ to hold. The intuitive reasoning behind this is that the eigenvalues of a matrix contain no information about the extent to which a matrix has linearly dependent columns other than the fact that $\lambda = 0 \implies $ $T$ is singular. Singular values do quantify this information, so intuitively, the two can only be equivalent if the eigenvectors are "as linearly independent as possible", i.e. orthogonal. Obviously this is not very precise but it gives you the idea.
To give a concrete example, consider the matrix $$A = \begin{pmatrix}1 &1 \\ 0 & 0\end{pmatrix}.$$ This has eigenvalues $1$ and $0$, but has singular values $\sqrt{2}$ and $0$. Here this discrepency is because the eigenvalue is simply telling you how much $A$ scales the eigenvector $(1,0)^T$, whereas the singular value is giving a measure of how much of the "mass" of the matrix is concentrated in the corresponding rank 1 operator that projects onto $(0,1)^T$ via the singular vectors, i.e. $u_1v_1^T$. Here "mass" can be more accurately characterized by the Frobenius or induced 2-norm.