Evaluate Limits Involving Trigonometric and Logarithmic Functions

To find the value of $a+b+m$, we evaluate the given limit:

$\lim _{x \rightarrow 2} \frac{\sin \left(x^3-5 x^2+a x+b\right)}{(\sqrt{x-1}-1) \log _e(x-1)}=m$

Let $x = 2 + h$. As $x \to 2$, $h \to 0$.

Step 1: Analyze the Denominator

The denominator becomes: $D(h) = (\sqrt{1+h}-1) \log_e(1+h)$

Using standard expansions as $h \to 0$: $\sqrt{1+h} - 1 = \frac{h}{\sqrt{1+h} + 1} = \frac{h}{2} + O(h^2)$ $\log_e(1+h) = h + O(h^2)$

Therefore, the denominator behaves as: $D(h) = \left(\frac{h}{2} + O(h^2)\right)\left(h + O(h^2)\right) = \frac{h^2}{2} + O(h^3)$

Since $D(h) \to 0$ as $h \to 0$, for the limit $m$ to exist and be finite, the numerator must also approach $0$.

Step 2: Determine $a$ and $b$

Let $P(x) = x^3 - 5x^2 + ax + b$. For $\sin(P(x)) \to 0$ as $x \to 2$, we require: $P(2) = 0$

Substitute $x = 2$: $P(2) = 2^3 - 5(2)^2 + 2a + b = 8 - 20 + 2a + b = 2a + b - 12 = 0 \implies 2a + b = 12 \quad \text{--- (1)}$

Since the denominator is of order $h^2$, for the limit to be finite, $P(2+h)$ must also be of order $h^2$. Writing the Taylor expansion of $P(2+h)$ around $h=0$: $P(2+h) = P(2) + P'(2)h + \frac{P''(2)}{2!}h^2 + \dots$

Since $P(2) = 0$, we must also have the coefficient of $h$ equal to $0$ (i.e., $P'(2) = 0$).

Differentiating $P(x)$: $P'(x) = 3x^2 - 10x + a$

Evaluating at $x = 2$: $P'(2) = 3(2)^2 - 10(2) + a = 12 - 20 + a = a - 8 = 0 \implies a = 8$

Substitute $a = 8$ into equation (1): $2(8) + b = 12 \implies 16 + b = 12 \implies b = -4$

Step 3: Evaluate the Limit $m$

With $a = 8$ and $b = -4$, the polynomial is: $P(x) = x^3 - 5x^2 + 8x - 4 = (x-2)^2(x-1)$

In terms of $h = x - 2$: $P(2+h) = h^2(2+h-1) = h^2(1+h)$

Now we substitute this back into the limit: $m = \lim_{h \to 0} \frac{\sin\left(h^2(1+h)\right)}{(\sqrt{1+h}-1) \log_e(1+h)}$

We rewrite the expression as: $m = \lim_{h \to 0} \left( \frac{\sin\left(h^2(1+h)\right)}{h^2(1+h)} \right) \cdot \left( \frac{h^2(1+h)}{(\sqrt{1+h}-1) \log_e(1+h)} \right)$

Using the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$: $m = 1 \cdot \lim_{h \to 0} \frac{1+h}{\left(\frac{\sqrt{1+h}-1}{h}\right) \left(\frac{\log_e(1+h)}{h}\right)}$

Applying the standard limits: $\lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} = \frac{1}{2}$ $\lim_{h \to 0} \frac{\log_e(1+h)}{h} = 1$

Thus: $m = \frac{1+0}{\frac{1}{2} \cdot 1} = 2$

Step 4: Calculate $a+b+m$

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