If _ x → 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a +...

Let $x - 2 = t$. As $x \to 2$, $t \to 0$.

The given limit can be written as:

$\lim_{t \to 0} \dfrac{\sin((t+2)^3 - 5(t+2)^2 + a(t+2) + b)}{(\sqrt{t+1} - 1)\log_e(t+1)} = m$

The denominator can be approximated for small $t$ as:

$\lim_{t \to 0} \dfrac{\sqrt{t+1} - 1}{t} \cdot \dfrac{\log_e(t+1)}{t} \cdot t^2 = \dfrac{1}{2} \cdot 1 \cdot t^2 = \dfrac{t^2}{2}$

For the limit to be finite, the argument of the sine function in the numerator must have $t^2$ as its lowest degree term in its expansion.

Let $P(t) = (t+2)^3 - 5(t+2)^2 + a(t+2) + b$

$P(t) = t^3 + 6t^2 + 12t + 8 - 5(t^2 + 4t + 4) + at + 2a + b$

$P(t) = t^3 + t^2 + (a - 8)t + (2a + b - 12)$

For the limit to exist, the constant and linear terms must be zero:

$a - 8 = 0 \Rightarrow a = 8$

$2a + b - 12 = 0 \Rightarrow 16 + b - 12 = 0 \Rightarrow b = -4$

Substituting $a$ and $b$ back into $P(t)$:

$P(t) = t^3 + t^2$

The limit becomes:

$m = \lim_{t \to 0} \dfrac{\sin(t^3 + t^2)}{t^2 / 2}$

$m = \lim_{t \to 0} \dfrac{\sin(t^2(t+1))}{t^2(t+1)} \cdot \dfrac{t^2(t+1)}{t^2 / 2} = 1 \cdot 2 = 2$

Therefore, $a + b + m = 8 - 4 + 2 = 6$.

Answer: $6$