$f(3)=\lim\limits_{x\to 3^-}f(x)=\lim\limits_{x\to 3^-}\dfrac{m}{3}(x+m+1)=\dfrac{m}{3}(m+4)=\dfrac{m^2}{3}+\dfrac{4m}{3}$
$\lim\limits_{x\to 3^+}f(x)=\lim\limits_{x\to 3^+}\dfrac{3-x}{5-\sqrt{x^2+16}}=\lim\limits_{x\to 3^+}\dfrac{(3-x)(5+\sqrt{x^2+16}) }{25-x^2-16}=\lim\limits_{x\to 3^+}\dfrac{(3-x)(5+\sqrt{x^2+16})}{(3-x)(3+x)}=\lim\limits_{x\to 3^+}\dfrac{5+\sqrt{x^2+16}}{3+x}=\dfrac{5+5}{3+3}=\dfrac{5}{3}$
Để $f(x)$ liên tục tại $x_o=3$:
$\dfrac{m^2}{3}+\dfrac{4m}{3}-\dfrac{5}{3}=0$
$\Leftrightarrow m=1$ hoặc $m=-5$
Chọn $A$
