$\because$ Denominator $=0$ at $x=2$

$\Rightarrow$ Numerator $=0$ at $x=2$

$\Rightarrow 2^3-5(2{)}^2+a(2)+b=0$

$\Rightarrow 2a+b=12 \qquad \cdots (1)$

$\therefore\; m= \lim_{x\to 2} \dfrac{ \dfrac{\sin(x^3-5x^2+ax+b)} {(x^3-5x^2+ax+b)} \cdot (x^3-5x^2+ax+b) } { \left( \dfrac{x-1-1} {\sqrt{x-1}+1} \right) \cdot \dfrac{\log_e(1+(x-2))} {(x-2)} \cdot (x-2) } $

$\Rightarrow m= \lim_{x\to 2} 2\left( \dfrac{x^3-5x^2+ax+b} {(x-2)^2} \right)$

$\Rightarrow m= \lim_{x\to 2} 2\left( \dfrac{3x^2-10x+a} {2(x-2)} \right)$

$\because$ Denominator $=0$ at $x=2$

$\Rightarrow$ Numerator $=0$ at $x=2$

$\Rightarrow 3(2)^2-10(2)+a=0 \Rightarrow a=8$

Put $a=8$ in (1), we get

$b=-4$

$\therefore m= \lim_{x\to 2} 2\left( \dfrac{6x-10}{2} \right) \Rightarrow m=2$

$\therefore a+b+m=6$