To bound a determinant of a matrix from above it's quite common to apply Hadamard's inequality. Unfortunately, in the following problem Hadamard's inequality isn't good enough. Are there other methods to bound determinants from above?
For $t\in [0,1]$, let $A_t$ be an $(n\times n)$-matrix over the complex numbers. Assume that $A_t$ is an invertible matrix for $t>0$ and that $A_0 = A_{t=0}$ is a singular matrix. Moreover, we are given that the limit $ L=\lim_{t\to 0^+} \frac{\vert\det A_t\vert}{t}$ exists. Also, the limit has the following form. Write $A_t = (a_{ij,t})_{i,j=1}^n$. Thus, the $(ij)$-th entry of $A_t$ is $a_{ij,t}$. Then $a_{ij,0} = a_i$, where $a_i$ is a non-zero complex number. (So the $i$-th row of $A_0$ is $(a_i,a_i,\ldots,a_i)$.)
For example, if $n=1$, we have that $A_t$ is a function going to zero.
If $n=2$, we can take the matrix $A_t=\left( \begin{array}{cc} 1+\sqrt{t} & 1+2\sqrt{t} \\ 2+\sqrt{t} & 2+2\sqrt{t}\end{array}\right).$ Note that $A_t$ converges to a singular matrix and that $A_t$ is invertible for any $0
Goal. Bound $L$ from above.
Unfortunately, Hadamard's inequality is useless in this case. In fact, Hadamard's inequality states that $\vert \det A_t\vert $ is bounded from above by a certain number. This certain number does not go to $0$ as $t\to 0$. Therefore, Hadamard's inequality gives the trivial upper bound $L \leq \infty$.
Therefore I am led to ask you.
Question. Are there other ways one can obtain upper bounds on absolute values of determinants?
Remark. Note that the example given by user3296 below $A_t = \lambda I_{n} t$ does not fulfill the conditions of my $A_t$. His matrix converges to the zero matrix.