Cho hàm số \(f(x) = \left\{ \begin{array}{l}\frac{{{x^3} - 4{x^2} + 3x}}{{\sqrt {x - 3} }}{\rm{              &n

Ta có: \(f(3)=0\)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{{x^3} - 4{x^2} + 3x}}{{\sqrt {x - 3} }} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{(x - 3)({x^2} - x)\sqrt {x - 3} }}{{x - 3}}\\
                  = \mathop {\lim }\limits_{x \to {3^ + }} ({x^2} - x)\sqrt {x - 3}  = 0
\end{array}\)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} - (m + 3)x + 3m{\rm{ }}}}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{{\rm{(x - 3)(x - m) }}}}{{x - 3}}\\
                  = \mathop {\lim }\limits_{x \to {3^ - }} {\rm{(x - m) = 3 -  m}}
\end{array}\)

Hàm số liên tục tại x = 1 khi và chỉ khi \(3 - m = 0 \Leftrightarrow m = 3\)