What is $\frac{\det(A+tI)}{\det(B+tI)}$ as $t\to0$? If $A$ and $B$ are two real $2\times 2$ matrices with $\det A = 0 $ and $\det B = 0 $ and $\mathrm{tr}(B)$ is non zero. then what will be limit of $$\lim_{t\to0}\frac{\det(A+tI)}{\det(B+tI)}$$ 
I used the formula $\lambda^2-\mathrm{tr} A+\det A = 0$. then i think answer is $\dfrac{\mathrm{tr}(A)}{\mathrm{tr}(B)}$. Am I correct?
What would be expansion of $\det(A+tI)$ for a $2\times 2$ matrices?
 A: Let $A$ be an $n \times n$ matrix. Then, one may check that 
$$\det (A + t I) = t^n + t^{n-1} \operatorname{tr} A + \cdots + t \, \lambda(A) + \det A$$
where $\lambda (A)$ is a certain constant depending on the eigenvalues of the matrix $A$. In the case $n = 2$, $\lambda (A) = \operatorname{tr} A$. So, if $A$ and $B$ are both $n \times n$ matrices,
$$\frac{\det (A + t I)}{\det (B + t I)} = \frac{t^n + t^{n-1} \operatorname{tr} A + \cdots + t \, \lambda(A) + \det A}{t^n + t^{n-1} \operatorname{tr} B + \cdots + t \, \lambda(B) + \det B}$$
Thus, if $\det A = \det B = 0$, and $\lambda (B) \ne 0$,
$$\lim_{t \to 0} \frac{\det (A + t I)}{\det (B + t I)} = \frac{\lambda (A)}{\lambda (B)}$$
and in the special case $n = 2$ this reduces to exactly what you guessed.
A: $
\def\a{\alpha}\def\b{\beta}
\def\z{0}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big(#1\Big)}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\adj#1{\operatorname{adj}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
\def\A{\trace{A^{-1}}}
\def\B{\trace{B^{-1}}}
$Define the matrix variables
$$\eqalign{
A &= \LR{A_0+tI} \qiq \dot A = I \\
B &= \LR{B_0+tI} \qiq \dot B = I \\
}$$
Jacobi's formula tells us that
$$\eqalign{
&\a = \det(A) \qiq\dot\a
    = \a\,\trace{A^{-1}\dot A} = \trace{\adj A} \\
&\b = \det(B) \qiq\dot\b
    = \b\,\trace{B^{-1}\dot B}  = \trace{\adj B} \\
}$$
Applying l'Hopital's Rule to your problem yields
$$\eqalign{
\lim_{t\to\z}\fracLR{\a}{\b}
 = \lim_{t\to\z}\fracLR{\dot\a}{\dot\b}
 = \frac{\trace{\adj{A_0}}}{\trace{\adj{B_0}}}
\\
}$$
For a $2\times 2$ matrix, the adjugate has a simple algebraic form
$$\eqalign{
\adj A &= I_2\trace A - A \\
}$$
Taking the trace of both sides yields
$$\eqalign{
\trace{\adj{A}} \,=\, \BR{\trace{I_2} - {\tt1}}\,{\trace{A}}
     \;\equiv\; \trace{A} \\
}$$
which confirms your conclusion.
