# Doubt in continuity

#1

#2

try to diffrenciate determinent u will get constant polynomial hence can be divisivle by fx..

#3

Since \alpha is the only root of f(x) it suffices to show that \alpha is also a double root of the given determinant

g(x) =\begin{vmatrix} A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha)\\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{vmatrix}

We can see that g(\alpha)=0 as rows 1 and 2 become identical.

Now g'(x) = \begin{vmatrix} A'(x) & B'(x) & C'(x) \\ A(\alpha) & B(\alpha) & C(\alpha)\\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{vmatrix}

g'(\alpha)=0 as 1st and 3rd rows become identical.

Since g(x) is a polynomial, this means \alpha is a double root of g(x) and hence divisible by f(x)