If $$\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1} - 1) \log_e(x-1)} = m$$, then $$a + b + m$$ is equal to :

Let the denominator be $$D(x) = (\sqrt{x-1} - 1) \ln(x-1)$$.

As $$x \to 2$$:

• $$\sqrt{2-1} - 1 = 1 - 1 = 0$$
• $$\ln(2-1) = \ln(1) = 0$$

Since the denominator approaches $$0$$, the numerator must also approach $$0$$ for the limit to be finite.

The numerator is $$N(x) = \sin(x^3 - 5x^2 + ax + b)$$.

For $$\lim_{x \to 2} N(x) = 0$$, the argument of the sine function must be $$0$$:

$$2^3 - 5(2^2) + a(2) + b = 0$$  $$8 - 20 + 2a + b = 0 \implies 2a + b = 12 \quad \text{--- (Eq. 1)}$$

Furthermore, since the denominator has two factors that both go to zero, the numerator's argument must contain $$(x-2)^2$$ as a factor to cancel out the "intensity" of the zero in the denominator. Let $$f(x) = x^3 - 5x^2 + ax + b$$. For $$(x-2)^2$$ to be a factor, $$f(2) = 0$$ and $$f'(2) = 0$$.

$$f'(x) = 3x^2 - 10x + a$$ $$f'(2) = 3(4) - 10(2) + a = 0 \implies 12 - 20 + a = 0 \implies \mathbf{a = 8}$$

Substitute $$a = 8$$ into (Eq. 1):

$$2(8) + b = 12 \implies 16 + b = 12 \implies \mathbf{b = -4}$$

Now we evaluate the limit with $$a=8, b=-4$$:

$$\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + 8x - 4)}{(\sqrt{x-1}-1) \ln(1+(x-2))}$$

Since $$\sin \theta \approx \theta$$ and $$\ln(1+\theta) \approx \theta$$ as $$\theta \to 0$$:

$$m = \lim_{x \to 2} \frac{x^3 - 5x^2 + 8x - 4}{(\sqrt{x-1}-1)(x-2)}$$

Factor the numerator: $$x^3 - 5x^2 + 8x - 4 = (x-2)^2(x-1)$$.

Rationalize the denominator part: $$(\sqrt{x-1}-1) = \frac{(x-1)-1}{\sqrt{x-1}+1} = \frac{x-2}{\sqrt{x-1}+1}$$.

Substitute these:

$$m = \lim_{x \to 2} \frac{(x-2)^2(x-1)}{\frac{x-2}{\sqrt{x-1}+1} \cdot (x-2)} = \lim_{x \to 2} (x-1)(\sqrt{x-1}+1)$$

$$m = (2-1)(\sqrt{2-1}+1) = 1(1+1) = \mathbf{2}$$

The question asks for $a + b + m$:

$$a + b + m = 8 + (-4) + 2 = 6$$

Correct Option: B (6)